Problem: Let LCP(s, t) be the length of the longest common prefix of strings s and t. Also let s[x ... y] be the substring of s from index x to index y (inclusive). For example, if s =  "abcde", then s[1 ... 3] = "abc", s[2 ... 5] = "bcde".

You are given a string s of length n and q queries. Each query is a pair of integer sets a_1, a_2, ..., a_k and b_1, b_2, ..., b_l. Calculate ∑_{i = 1}^{i = k} ∑_{j = 1}^{j = l}{LCP(s[a_i ... n], s[b_j ... n])} for each query.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 2 ⋅ 10^5) — the length of string s and the number of queries, respectively.

The second line contains a string s consisting of lowercase Latin letters (|s| = n).

Next 3q lines contains descriptions of queries — three lines per query. The first line of each query contains two integers k_i and l_i (1 ≤ k_i, l_i ≤ n) — sizes of sets a and b respectively.

The second line of each query contains k_i integers a_1, a_2, ... a_{k_i} (1 ≤ a_1 < a_2 < ... < a_{k_i} ≤ n) — set a.

The third line of each query contains l_i integers b_1, b_2, ... b_{l_i} (1 ≤ b_1 < b_2 < ... < b_{l_i} ≤ n) — set b.

It is guaranteed that ∑_{i = 1}^{i = q}{k_i} ≤ 2 ⋅ 10^5 and ∑_{i = 1}^{i = q}{l_i} ≤ 2 ⋅ 10^5.

Output

Print q integers — answers for the queries in the same order queries are given in the input.

Example

Input

7 4
abacaba
2 2
1 2
1 2
3 1
1 2 3
7
1 7
1
1 2 3 4 5 6 7
2 2
1 5
1 5


Output

13
2
12
16

Note

Description of queries: 

  1. In the first query s[1 ... 7] = abacaba and s[2 ... 7] = bacaba are considered. The answer for the query is LCP(abacaba, abacaba) + LCP(abacaba, bacaba) + LCP(bacaba, abacaba) + LCP(bacaba, bacaba) = 7 + 0 + 0 + 6 = 13.
  2. In the second query s[1 ... 7] = abacaba, s[2 ... 7] = bacaba, s[3 ... 7] = acaba and s[7 ... 7] = a are considered. The answer for the query is LCP(abacaba, a) + LCP(bacaba, a) + LCP(acaba, a) = 1 + 0 + 1 = 2.
  3. In the third query s[1 ... 7] = abacaba are compared with all suffixes. The answer is the sum of non-zero values: LCP(abacaba, abacaba) + LCP(abacaba, acaba) + LCP(abacaba, aba) + LCP(abacaba, a) = 7 + 1 + 3 + 1 = 12.
  4. In the fourth query s[1 ... 7] = abacaba and s[5 ... 7] = aba are considered. The answer for the query is LCP(abacaba, abacaba) + LCP(abacaba, aba) + LCP(aba, abacaba) + LCP(aba, aba) = 7 + 3 + 3 + 3 = 16. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int edges, n, Q;
int hd[400004], to[400004], nex[400004], tr[400004];
char str[400004];
void addedge(int u, int v, int c) {
  nex[++edges] = hd[u], hd[u] = edges, to[edges] = v;
}
namespace SAM {
int last, tot;
int len[400004], f[400004], trans[400004][27];
void init() { last = tot = 1; }
int extend(int c) {
  int np = ++tot, p = last;
  len[np] = len[p] + 1, last = np;
  while (p && !trans[p][c]) trans[p][c] = np, p = f[p];
  if (!p)
    f[np] = 1;
  else {
    int q = trans[p][c];
    if (len[q] == len[p] + 1)
      f[np] = q;
    else {
      int nq = ++tot;
      len[nq] = len[p] + 1;
      memcpy(trans[nq], trans[q], sizeof(trans[q]));
      f[nq] = f[q], f[q] = f[np] = nq;
      while (p && trans[p][c] == q) trans[p][c] = nq, p = f[p];
    }
  }
  return np;
}
void build() {
  for (int i = 2; i <= tot; ++i) addedge(f[i], i, len[f[i]] - len[i]);
}
}  // namespace SAM
int tim;
int dfn[400004], Top[400004], hson[400004], siz[400004], fa[400004],
    dis[400004];
void dfs1(int u, int ff) {
  fa[u] = ff, siz[u] = 1, dfn[u] = ++tim, dis[u] = dis[ff] + 1;
  for (int i = hd[u]; i; i = nex[i]) {
    int v = to[i];
    if (v == ff) continue;
    dfs1(v, u);
    siz[u] += siz[v];
    if (siz[v] > siz[hson[u]]) hson[u] = v;
  }
}
void dfs2(int u, int tp) {
  Top[u] = tp;
  if (hson[u]) dfs2(hson[u], tp);
  for (int i = hd[u]; i; i = nex[i]) {
    int v = to[i];
    if (v == fa[u] || v == hson[u]) continue;
    dfs2(v, v);
  }
}
int LCA(int x, int y) {
  while (Top[x] != Top[y]) {
    dis[Top[x]] > dis[Top[y]] ? x = fa[Top[x]] : y = fa[Top[y]];
  }
  return dis[x] < dis[y] ? x : y;
}
vector<int> G[400004];
long long ans = 0;
int t = 0, tot = 0;
int arr[400004], brr[400004], A[400004], mk[400004], mka[400004], mkb[400004],
    sizea[400004], sizeb[400004], S[400004];
bool cmp(int i, int j) { return dfn[i] < dfn[j]; }
void addvir(int x, int y) { G[x].push_back(y); }
void insert(int x) {
  if (t <= 1) {
    S[++t] = x;
    return;
  }
  int lca = LCA(S[t], x);
  if (lca == S[t]) {
    S[++t] = x;
    return;
  }
  while (t > 1 && dis[S[t - 1]] >= dis[lca]) addvir(S[t - 1], S[t]), --t;
  if (S[t] != lca) addvir(lca, S[t]), S[t] = lca;
  S[++t] = x;
}
void DP(int x) {
  for (int i = 0; i < G[x].size(); ++i)
    DP(G[x][i]), sizea[x] += sizea[G[x][i]], sizeb[x] += sizeb[G[x][i]];
  for (int i = 0; i < G[x].size(); ++i) {
    ans += 1ll * (sizea[x] - sizea[G[x][i]]) * sizeb[G[x][i]] * SAM::len[x];
  }
  ans += (1ll * mka[x] * mkb[x] + 1ll * mka[x] * sizeb[x] +
          1ll * mkb[x] * sizea[x]) *
         SAM::len[x];
  sizea[x] += mka[x], sizeb[x] += mkb[x];
}
void erase(int x) {
  mka[x] = mkb[x] = sizea[x] = sizeb[x] = 0;
  for (int i = 0; i < G[x].size(); ++i) erase(G[x][i]);
  G[x].clear();
}
void work() {
  tot = 0;
  int k1, k2;
  scanf("%d%d", &k1, &k2);
  for (int i = 1; i <= k1; ++i)
    scanf("%d", &arr[i]), A[++tot] = tr[arr[i]], mka[tr[arr[i]]] = 1;
  for (int i = 1; i <= k2; ++i)
    scanf("%d", &brr[i]), A[++tot] = tr[brr[i]], mkb[tr[brr[i]]] = 1;
  sort(A + 1, A + 1 + tot, cmp);
  tot = unique(&A[1], &A[tot + 1]) - A - 1;
  t = ans = 0;
  if (A[1] != 1) S[t = 1] = 1;
  for (int i = 1; i <= tot; ++i) insert(A[i]);
  while (t > 1) addvir(S[t - 1], S[t]), --t;
  DP(1);
  printf("%I64d\n", ans);
  erase(1);
}
int main() {
  scanf("%d%d%s", &n, &Q, str + 1);
  SAM::init();
  for (int i = n; i >= 1; --i) tr[i] = SAM::extend(str[i] - 'a');
  SAM::build();
  dfs1(1, 0), dfs2(1, 1);
  for (int i = 1; i <= Q; ++i) work();
  return 0;
}