Introduction
The research of above/below guarantee parameters as first used by Mahajan and Raman [22] was an important step towards studying problems whose natural parameters provided only trivial and unsatisfactory answers. Case in point, the motivation for Mahajan and Raman was the observation that every CNFSAT formula with clauses trivially has an assignment that satisfies clauses, thus question for the maximum number of satisfied clauses is only interesting if , which of course renders the parametrised approach unnecessary. They therefore proposed to study parametrisations ‘above guarantee’: going with the previous example, we would ask to satisfy clauses or ‘ above guarantee’. After some isolated results in that direction ( [15, 17]) the programme took up steam after Mahajan et alpresented several results and pointers in new directions [23] ( [3, 16, 18, 17]). In particular, Cygan et albroke new ground for Multiway Cut and Vertex Cover with algorithms that run in time, where is the gap parameter between an appropriate LPrelaxation and the integral optimum [5]. Lokshtanov et alimproved the Vertex Cover case to using a specialized branching algorithm [21].
The latter result highlights an important realization: these alternative, smaller parameters might not only provide the means to investigate problems without ‘good’ natural parameters, it might also provide us with faster algorithms in practise! Gap and aboveguarantee parameters are attractive because there is a reasonable chance that they are small in realworld scenarios, something we often cannot expect from natural parameters.
To the best of our knowledge, so far no gapparameter results are known for domination problems and an above/belowguarantee result is only known in boundeddegree graphs [23]
. This is probably due to the fact that there are no simple ‘natural’ upper/lower bounds and in the case of gapparameters the LPdualities do not provide much purchase. We therefore explore this topic under the most basic assumptions: we are provided with a witness for a lower bound on the domination number
as input and consider parametrisations that arise from this additional information. In the case of prob]Dominating Set@Dominating SetDominating Set, the witness takes the form of a independent set, that is, a set in which all vertices have pairwise distance at least three. Note that this approach also captures a form of duality: the LPdual of dominating set describes a independent set, however, in general the two optima are arbitrarily far apart. Recently, Dvořák highlighted this connection [7] and proved that in certain sparse classes the gap between the dual optima is bounded by a constant.Thus, assume we are given a maximal independent set alongside the input graph . A parametrisation by would go against the spirit of gapparameters, instead we parametrise by the size of residual set , that is, all vertices that lie at distance two from (since is maximal, no vertex can have distance three or more). We choose this particular parameter for two reasons:

For the problem is decidable in polynomial time since the domination number of the graph is precisely .

The set is a dominating set of .
The first property is of course an important prerequisite for the problem to be in under this parametrisation, while the second property guarantees us that the dominating set size lies inbetween and .
Our first investigatory dimension is the constant in the independent set: intuitively, increasing the minimum distance between vertices in increases the size of the parameter and imposes more structure on the input instance. Our second dimension encompasses an approach that has been highly successful in improving tractability of domination problems: restricting the inputs to sparse graphs. While prob]Dominating Set@Dominating SetDominating Set is complete in general graphs, Alber et alshowed that it is fpt in planar graphs [1]; Alon and Gutner later proved that assuming degeneracy is sufficient [4]. Philip, Raman, and Sikdar extended this result yet further to graphs excluding a fixed biclique and also proved that it admits a polynomial kernel [25]. A related line of research was the hunt for linear kernels in sparse classes. Beginning with such a kernel on planar graphs by Alber, Fellows, and Niedermeier [2], results on apexminor free graphs [10], graphs excluding a minor [11] and classes excluding a topological minor [12] were soon proven. Recently, a linear kernel for graphs of bounded expansion [6] (and an almostlinear kernel for nowhere dense graphs [9]) has subsumed all previous results.
Our investigation of prob]Dominating Set@Dominating SetDominating Set parametrised above an independent set, for , led us to the following results. For , the problem is complete already for , squashing all hope for an or even algorithm. This also holds true if the inputs are restricted to sparse graph classes (apexgraphs/graphs of maximum degree six).
For , the problem is hard in general graphs but admits an algorithm. In nowhere dense and bounded expansion classes, it is fixedparameter tractable. We further show, in the probably most technical part of this paper, that it admits a linear kernel in bounded expansion classes.
Finally, for the problem remains hard in general graphs and essentially degenerates to prob]Dominating Set@Dominating SetDominating Set (hence, all the above mentioned results in sparse classes translate in the parametrisation above independence).
1 Preliminaries
A set is independent if each pair of distinct vertices in have distance at least , thus an independent set is independent. We write and , respectively, for the neighbourhood and the closed neighbourhood of a vertex . We extend this notation to sets as follows: for we let be all vertices not in that have a neighbour in and . We let be all vertices not in that are at most distance from any vertex in and we let . A vertex set is dominated by a set if for every vertex we have , is then called a dominator. We let denote the size of a minimum dominating set of .
Note that is trivially a dominating set and that, for , it holds that , thus we will tacitly assume in the following that since all other instances are trivial.
We will frequently invoke the terms bounded expansion and nowhere dense to describe graph classes. The definitions of these terms requires the introduction of several concepts which will not be useful for the remainder of the paper, we refer the reader to the book by Nešetřil and Ossona de Mendez [24]. In this context, it is important to know that bounded expansion classes generalize most structurally sparse classes (planar, bounded genus, bounded degree, minor free, topological minor free) and nowhere dense classes contain bounded expansion classes in turn. The following lemma and propositions for those two sparse graph classes will be needed in the remainder of this paper:
[Twin class lemma [13, 26]] For every bipartite nowhere dense class there exists a constant and a function such that for every member of the class it holds that

, and

.
where . If is from a bounded expansion class, can be assumed a constant as well.
We will also frequently invoke the following result regarding firstorder (FO) model checking in bounded expansion and nowhere dense classes:
Proposition (Dvořák, Král, and Thomas [8]).
For every bounded expansion class, the firstorder model checking problem is solvable in linear fpttime parametrised by the size of the input formula.
This result has since been extended to nowhere dense classes as well. Here, almost linear fpttime means running time of the form for some function .
Proposition (Grohe, Kreutzer, and Siebertz [14]).
For every nowhere dense class, he firstorder model checking problem is solvable in almost linear fpttime parametrised by the size of the input formula.
2 Above independence: hard as nails
In this section we will show that when we let , we find that the problem is complete for , hence this parametrisation does not even admit an algorithm. In the following we first present a reduction from prob]3SAT@3SAT3SAT and then discuss how to modify it to reduce into sparse graph classes.
Since is a (maximal) independent set, we know that each vertex in is a neighbour of some vertex in , otherwise we could add this vertex to . Let us now describe the reduction. Let be instance with variables and clauses . We construct as follows (Figure 1):

For each variable , add a triangle with vertices , , .

For each clause add a vertex . If the variable occurs positively in , add the edge ; if it occurs negatively, add the edge .

Add a single vertex to the graph and connect it to each clause variable . Add two further vertices and add the edges and .
We further set as our independent set; notice that the only vertex not contained in is . Hence, .
is satisfiable iff has a dominating set of size .
Proof.
Assume is satisfiable and fix one satisfying assignment . We construct a dominating set as follows: if is true under , add to ; otherwise add . Since satisfies every clause of the dominating set so far dominates every clause vertex and, of course, every variable gadget. The remaining undominated vertices are , thus adding to yields a dominating set of of size .
In the other direction, assume that is a dominating set of of size . Since is a pendant vertex we can assume that (if would be in we could exchange it for ). That leaves vertices in , precisely the number of variablegadgets. Since every variable gadget must include at least one vertex of , we conclude the every such gadget contains precisely one dominating vertex. Since that depletes our budget, no other vertex is contained in .
By the usual exchange argument we may assume that the dominating vertex in each variable gadget is either or and not for ; hence the dominating vertices inside the variable gadgets encode a variable assignment of . Finally, note that the clause vertices are not dominated by and is not contained in . Hence, they must be completely dominated by vertices contained in the variable gadgets. Then, by construction, the assignment satisfies and the claim follows. ∎
We conclude that manyone reduces to prob]Dominating Set@Dominating SetDominating Set above independence already with . We obtain the following two corollaries that demonstrate that sparseness cannot help tractability here:
prob]Dominating Set@Dominating SetDominating Set above independence is complete in apexgraphs.
Proof.
We use the above construction but reduce from a planar variant of prob]3SAT@3SAT3SAT. To ensure that we can construct variablegadgets without edge crossings, we choose to reduce from Lichtenstein’s prob]Planar 3SAT@Planar 3SATPlanar 3SAT variant [20] which ensures that the following graph derived from the prob]Planar 3SAT@Planar 3SATPlanar 3SAT instance is planar:

Every variable of is represented by two literal vertices with the edge

Each clause is represented by a vertex . If the variable occurs positively in , the edge exits; if it occurs negatively, the edge exists.
To complete to we have to add the vertices and connect them to . This is clearly possible without breaking planarity (picture placing on the middle of the line segment and moving it perpendicular by a small amount, then the edges and can be embedded without crossing other edges). The vertices can be placed anywhere; finally the vertex will break planarity (the embedding does not guarantee that the clause vertices lie on the outer face of the graph) and we conclude that is indeed an apexgraph. ∎
prob]Dominating Set@Dominating SetDominating Set above independence is complete in graphs of maximum degree six.
Proof.
We reduce from prob](3,4)SAT@(3,4)SAT(3,4)SAT (hardness shown in [27]) in which every clause has size three and every variable occurs in at most four clauses. We use the above construction with one modification. Instead of connecting all clause vertices to one vertex we create a boundeddegree tree with the clausevertices as its leaves.
We begin by partitioning the clause vertices in pairs
; if there is an odd number of vertices the last group will have three vertices. Then, for each group
, we add two vertices , connect the clausevertices and to , and add the edge . We further add each to our independent set and then create a set consisting of each . Now we iteratively construct the next level of the tree, starting with :
If , create a single vertex , connect it to and finish, otherwise proceed with the next step.

Partition the vertices in into groups of pairs. If is odd, the last group will be a triple instead.

For each group, create a treegadget , with vertices (and if the group contains a third vertex), and edges , (), , , (), and .

Add each to , let now be the set of all (for ) and continue with Step 1.
Figure 1, on the right side, shows an example of this construction. We note that, in the last treegadget, and are the same vertices as and respectively in the figure. We conclude the construction by adding each from the variablegadgets to and setting . Notice that the only vertex not contained in is and thus .
Since each variable in prob](3,4)SAT@(3,4)SAT(3,4)SAT can be in up to four clauses, the maximum degree for and is six. All clausevertices have degree at most four and all other types of vertices have a degree not higher than that, hence the claimed degreebound holds. It is left to show that is satisfiable iff there is a dominating set of size in the graph.
Let us assume that is satisfiable and fix one satisfying assignment . We construct a dominating set as follows, beginning in the same way as in Lemma 2: if is true under , add to , otherwise add . Since satisfies every clause of the dominating set so far dominates every clause vertex and every variable gadget. Now, the remaining undominated vertices are the treevertices, and our remaining budget is which is equal to the amount of treegadgets. Since every clause vertex is already dominated we can, for each treegadget , add to the dominating set. This will dominate and the corresponding or in the treegadget below it, hence we can dominate all vertices of the graph within the budget .
In the other direction, assume that is a dominating set of of size . Since is a pendant vertex we can assume that . Thus, in the last treegadget () is in the dominating set. This means that in order for and to be dominated, in both treegadgets above has to be in the dominating set. This holds for all treegadgets, all the way up to the clause vertices. Since we now have one vertex per treegadget in the dominating set this leaves vertices. Just as in Lemma 2, we note that, for each variable gadget, either or is in the dominating set. We know that the clause variables are not dominated by anything in the tree gadgets and thus must be dominated by the variable vertices. As stated in Lemma 2, the dominating vertices inside the variable gadgets encode a variable assignment that satisfies and the claim follows. ∎
3 Above independence: sparseness matters
3.1 hardness in general graphs
In the following we present a result for prob]Dominating Set@Dominating SetDominating Set above independence, namely that it is hard in general graphs. We show this by reduction from prob]Colourful Dominating Set@Colourful Dominating SetColourful Dominating Set parametrised by the number of colours :
It is easy to verify that prob]Colourful Dominating Set@Colourful Dominating SetColourful Dominating Set is hard by reducing from prob]RedBlue Dominating Set@RedBlue Dominating SetRedBlue Dominating Set: we copy the blue set times and make each copy a colour set , , and let be the red set.
prob]Dominating Set@Dominating SetDominating Set above independence is hard for .
Proof.
Let be an instance of prob]Colourful Dominating Set@Colourful Dominating SetColourful Dominating Set. We construct an instance of prob]Dominating Set@Dominating SetDominating Set above independence as follows (Figure 2):

Begin with equal to ; then

for each block , , add edges to make a complete graph and add an additional vertex with neighbourhood ; then

add a vertex and connect it to all vertices in .
Let be the independent set (which it clearly is for any ) and thus . Note that the graph trivially has a dominating set of size ; the set . We now claim that has a colourful dominating set of size iff has a dominating set of size .
Assume that has a solution of size and fix one such colourful dominating set . Note that in a) dominates all of (because it dominates in ) and b) dominates each set , (because intersects each such ), and of course the vertex . Hence, is a dominating set of as well.
In the other direction, assume that is a dominating set of of size . Since each , , is only connected to vertices in the corresponding set , at least one vertex from each set needs to be in . Since there are such sets we conclude that intersects each set in precisely one vertex. We can further modify any such solution to not take the vertices by taking an arbitrary vertex from instead, thus assume that has this form in the following. But then is of course also a dominating set of size for , as claimed.
∎
3.2 Tractability in sparse graphs
In the following we present two positive results, namely that prob]Dominating Set@Dominating SetDominating Set above independence is fpt in nowhere dense classes and that it admits a polynomial kernel in bounded expansion classes. The algorithm further implies an algorithm in general graphs. The following annotated domination problem will occur as a subproblem:
In the following algorithm we will group vertices of according to their neighbourhood in (or a subset of ). We will call those groups neighbourhood classes. We will write to denote the size of an optimal solution of prob]Annotated Dominating Set@Annotated Dominating SetAnnotated Dominating Set.
prob]Dominating set@Dominating setDominating set above independence can be solved in linear fpttime in any graph class of bounded expansion.
Proof.
First we guess the intersection of an optimal solution (should it exist) with in time. Let be those vertices of that are not dominated by . Define
as the neighbourhoods induced in by vertices in . By Lemma 1, we have that since is from a class with bounded expansion (note that the partition into such neighbourhoods is computable in linear time using the partitionrefinement data structure [19]). Accordingly, in time , we can guess a subset such that an optimal solution covers exactly the neighbourhoods (if does not cover we abort this branch of the computation). We are now left with the task of choosing vertices from to a) cover the neighbourhoods and b) dominate the vertices in which are not dominated by .
Let us introduce the following notation to ease our task: for a collection of neighbourhoods and a set , let . That is, contains those neighbourhood classes in whose neighbourhood is contained in . Let be an ordering of and let ; we will describe a dynamicprogramming algorithm over the ordering . Let be the minimum size of a partial solution in that covers the neighbourhoods and together with dominates all of . We initialize for all and , then compute the following entries with the recurrence^{1}^{1}1A proof for the correctness of the recurrence can be found in the Appendix
Note that is the minimum size of a set that dominates while choosing at least one vertex from each member of (if we assume that ). The latter constraint corresponds to dominating the neighbourhoods of in by using vertices from . Once the DP table has been computed, the size of an optimal solution is the value in .
It remains to be noted that every neighbourhood graph admits a dominating set of size one, hence the annotated dominating set has size at most . Thus, the problem of finding an annotated dominating set is FOexpressible by a formula of size and we can solve the subproblem of computing in time for some function using Proposition 1. As a result, we obtain a linear dependence on the input size (note that ) in the running time and thus the problem is solvable in linear fpttime.∎
prob]Dominating set@Dominating setDominating set above independence can be solved in almost linear fpttime in any nowhere dense class.
We finally note that the algorithm described in the proof of Lemma 3.2 only needs a blackbox fptalgorithm for prob]Annotated Dominating Set@Annotated Dominating SetAnnotated Dominating Set to run in fpt time, thus it is very likely that prob]Dominating set@Dominating setDominating set above independence is in for other highly structured but not necessarily sparse graph classes. We can run the same algorithm on general graphs to obtain an algorithm using the bound and a simple bruteforce in time on the prob]Annotated Dominating Set@Annotated Dominating SetAnnotated Dominating Set subinstance during the DP.
prob]Dominating set@Dominating setDominating set above independence is in .
3.3 Kernelization in sparse graphs
Let us now set up the necessary machinery for the kernelization. A boundaried graph is a tuple where is a graph and is the boundary. We also write to denote the boundary. For a graph and an induced subgraph , the boundary are those vertices of that have neighbours in . Thus for every subgraph of there is a naturally associated boundaried graph .
For a boundaried graph and subsets a set is an dominator of if and dominates the set . We let denote the size of a minimum dominator of . A replacement for is a boundaried graph with . The operation of replacing by in , written as , consist of removing the vertices from , then adding to with in identified with in (we assume that the vertices do not occur in ).
Let be an instance of prob]Dominating Set@Dominating SetDominating Set above independence where is from a bounded expansion class. Let , and . Then, in fpttime with parameter , we can compute a replacement for of size such that . Moreover, the replacement is a subgraph of and contains .
Proof.
For every pair of subsets we compute a minimal dominator for . Since this problem is expressible by an FOformula of size , we can employ Proposition 1 to compute the set in linear fpttime with parameter . Note that will, besides the vertices in , contain at most additional vertices, since vertices suffice to dominate and the vertex dominates all of .
Let be the union of all such computed solutions and the boundary. By construction, . Since all dominators live inside , we can safely remove the edges from that do not have any endpoint in , call the resulting graph . Let and let be a partition of into twin classes (thus all vertices in have the same neighbourhood in and no other class has this neighbourhood). Note that is an independent set in .
Let be obtained from by removing all but two representatives from every twinclass, denote those by . Clearly, every dominator of is still an dominator of , we need to proof the other direction. Let be an dominator of . If then is still an dominator for . The same holds if intersects every twinclass in at most one vertex: each such class either has size one, in which case it has size one in as well, or it has size two and the vertex not contained in is dominated by a vertex in . In either case, dominates all of and thus is an dominator of and therefore of . Thus, assume that fully contains some class . Clearly, only one vertex of is enough to dominate (and potentially vertices in ), thus we can modify by picking the central vertex instead to dominate the other vertex of (which of course also dominates all of in ). This can, of course, only happen once, otherwise we would reduce the size of the supposedly minimal set . This leads us back to the previous case and we conclude that there exists an dominator of and thus of equal size.
We conclude that for every choice of ; which implies that . Finally, by the twinclass lemma, since is a constant in bounded expansion classes. ∎
Let be an instance of prob]Dominating Set@Dominating SetDominating Set above independence where is from a bounded expansion class. Let be a subset and let be those vertices with . Let be the induced subgraph on . Assuming is not empty we can, in fpttime with parameter , compute a replacement for of size alongside an offset such that . Moreover, the replacement is a subgraph of .
Proof.
Let and define the graphs for . We first we apply Lemma 3.3 to replace every subgraph by a subgraph of size in linear fpttime with parameter . For simplicity, let us call the resulting graph and relabel the graphs to (note that Lemma 3.3 ensures that dominating set size does not change and that is still a independent set of the resulting graph).
For every we compute a
characteristic vector
indexed by pairs of subsets of with the following semantic: for we set . If is larger than we simply set .Note that the constraint subproblem to compute is FOexpressible, by a formula of size , thus we can compute the vectors for in linear time fpttime with parameter using Proposition 1. Let be the equivalence relation over the graphs defined as and let be the corresponding partition into equivalenceclasses under . Note that since that is the number of possible characteristic vectors.
The construction of is now simple: in every equivalenceclass we select subgraphs and remove the rest; clearly is a subgraph of of the claimed size. We let the offset to be equal to the number of subgraphs removed in this way.
We are left to show that . Consider any minimal dominating set for . We call a graph interesting under if intersects in any vertex besides .
Claim.
There exists a solution of size equal to under which at most graphs per equivalence class are interesting.
Proof.
Consider any such class . If we are done, so assume otherwise. Let be the set of vertices that are dominated through vertices in graphs contained in and select up to many graphs that together already dominate ; we let to be equivalent to on these graphs. Any graph not selected in this way only needs to dominate itself and we add its centre vertex to . Since every graph needs to intersect any dominating set in at least one vertex, and since is minimal we must have . Finally, only the selected graphs are interesting under , as claimed. ∎
Thus, let us assume in the following that is such a minimal solution under which at most graphs per class are interesting. For such a solution of , we construct a solution of of size as follows. For a class , let be those graphs that are contained in and let be the graphs that are interesting under . Since , we can pair every graph with a graph . Fix such a pair , let and let be those vertices of that are exclusively dominated vertices in . Since , there must exist a set of vertices in that dominate and . If we repeat this construction for every graph with their respective pair in and then pick the centre vertex for all , then the resulting set has size and dominates all of .
The same proof works in reverse if we start with a dominating set of to construct a dominating set of with ; thus we conclude that and the claim follows. ∎
prob]Dominating Set@Dominating SetDominating Set above independence has a linear kernel in bounded expansion graphs.
Proof.
Let be the input instance and let be the members of the independent set . We define the graphs and their respective neighbours for . Let be as in Lemma 1. We partition the graphs into two sets , where iff ; and contains all remaining graphs.
Let us first reduce . Let be the neighbourhoods of the graphs collected in . By the twin class lemma, , however, for each member we might have many graphs in that intersect in precisely this set .
Fix for now and let be those graphs of that intersect in . Let be the joint graph of the subgraphs in . Since , and is a constant depending only on the graph class, we can apply Lemma 3.3 to compute a replacement of size with offset in polynomial time. We apply the replacement and decrease by . Repeating this procedure for all yields a graph (a subgraph of ) in which the small graphs in total contain at most vertices, a independent set of , and a new input . This concludes our reduction for .
Let us now deal with in . By the twin class lemma, . But then has size bounded in and we conclude that the size of a minimal dominating set for is bounded by , hence we assume that is bounded by (otherwise the instance is positive and we can output a trivial instance). We now apply the existing linear kernel [6] for Dominating Set to . The output of the kernelization is a subgraph of the original graph, hence we collect the remaining vertices of the independent set in order to output a wellformed instance . This concludes the proof. ∎
4 Above independence: simple domination
In the case where the lowerbound set is independent we now have that for distinct it holds that , thus for each the set can only be dominated from and . Let us call an instance reduced if for every the intersection is nonempty. We can easily preprocess our input instance to enforce this property: if such an would exist we can simply remove from to obtain an equivalent instance. In a reduced instance the parameter is necessarily big compared to :
Observation .
For every reduced instance of prob]Dominating Set@Dominating SetDominating Set above independence it holds that .
Let be a graph class for which prob]Dominating Set@Dominating SetDominating Set is in . Then prob]Dominating Set@Dominating SetDominating Set above independence is in for as well.
Let be a graph class for which prob]Dominating Set@Dominating SetDominating Set admits a polynomial kernel. Then prob]Dominating Set@Dominating SetDominating Set above independence admits a polynomial kernel for as well.
On the other hand, we showed in Section 3.1 that the problem remains hard for any in general graphs.
5 Conclusion
We considered prob]Dominating Set@Dominating SetDominating Set parametrised by the residual of a given independent set and investigated how the value of and the choice of input graph classes affect its tractability. We observed that the tractability does improve from to as it goes from being complete to ‘merely‘ hard and at least admits an algorithm. Larger values of , however, do not increase the tractability as the problem becomes essentially equivalent to prob]Dominating Set@Dominating SetDominating Set.
If we consider sparse classes (bounded expansion and nowhere dense), the improvement in tractability from to is much more pronounced; changing from complete to and even admitting a linear kernel in bounded expansion classes. We very much believe that the kernel can be extended to nowhere dense classes, but leave that quite technical task as an open question.
Acknowledgments: We thank our anonymous reviewer for helpfully pointing out how to achieve a linear kernel in Theorem 3.3.
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Appendix
We claim that that is the size of a partial solution in that covers the neighbourhoods and together with dominates all of .
As the inductive base, we set for all and . Clearly, a nonempty cannot possibly be dominated (note that since we guessed , no vertex in is contained in the dominating set). The cost of dominating nothing in is of course , hence the choice of . We conclude that the tables of the base cases describe the claimed quantity.
Since the base cases hold, let us assume that the table for all describe the claimed partial solutions. Recall that the recurrence is given by
Let be a minimal set which covers and dominates all of . We argue in the following that .
For the first direction, let be the dominators of that lie outside of . Let then contain those sets of that are covered by vertices in and let be those that are covered by . Since covers all of , we know that and hence this pair appears in the above recurrence. By induction, we know that
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