As we all know, Aryan is a funny guy. He decides to create fun graphs. For a graph (G), he defines fun graph (G') of (G) as follows: Every vertex (v') of (G') maps to a non-empty independent set(^{\text{∗}}) or clique(^{\text{†}}) in (G). The sets of vertices of (G) that the vertices of (G') map to are pairwise disjoint and combined cover all the vertices of (G), i.e., the sets of vertices of (G) mapped by vertices of (G') form a partition of the vertex set of (G). If an edge connects two vertices (v_1') and (v_2') in (G'), then there is an edge between every vertex of (G) in the set mapped to (v_1') and every vertex of (G) in the set mapped to (v_2'). If an edge does not connect two vertices (v_1') and (v_2') in (G'), then there is not an edge between any vertex of (G) in the set mapped to (v_1') and any vertex of (G) in the set mapped to (v_2'). As we all know again, Harshith is a superb guy. He decides to use fun graphs to create his own superb graphs. For a graph (G), a fun graph (G' ') is called a superb graph of (G) if (G' ') has the minimum number of vertices among all possible fun graphs of (G). Aryan gives Harshith (k) simple undirected graphs(^{\text{‡}}) (G_1, G_2,\ldots,G_k), all on the same vertex set (V). Harshith then wonders if there exist (k) other graphs (H_1, H_2,\ldots,H_k), all on some other vertex set (V') such that: (G_i) is a superb graph of (H_i) for all (i\in \{1,2,\ldots,k\}). If a vertex (v\in V) maps to an independent set of size greater than (1) in one (G_i, H_i) ((1\leq i\leq k)) pair, then there exists no pair (G_j, H_j) ((1\leq j\leq k, j\neq i)) where (v) maps to a clique of size greater than (1). Help Harshith solve his wonder. (^{\text{∗}})For a graph (G), a subset (S) of vertices is called an independent set