// LLP-LIS: G[j] >= G[i] + 1 for every i in pre(j) (i < j with A[i] < A[j]).

class LLPLongestIncreasingSubseq {
  int[] LLPLongestIncreasingSubseq(int[] A, set<int>[] pre) {
    int[] G = 1;
    forbidden (j) :
      exists i in pre[j] : G[j] < G[i] + 1
    =>
      advance :
        G[j] = max i in pre[j] : G[i] + 1;
    return G;
  }
}