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;
const int N = 1e6 + 10;
char s[N];
int T[400010][30], cnt, fa[400010], len[400010], last;
int pos[N];
void build(int v) {
  int p = last, np, q, nq, i;
  np = last = ++cnt;
  len[np] = len[p] + 1;
  for (; p && T[p][v] == 0; p = fa[p]) T[p][v] = np;
  if (p == 0)
    fa[np] = 1;
  else {
    q = T[p][v];
    if (len[q] == len[p] + 1)
      fa[np] = q;
    else {
      nq = ++cnt;
      len[nq] = len[p] + 1;
      fa[nq] = fa[q];
      fa[q] = fa[np] = nq;
      for (; T[p][v] == q; p = fa[p]) T[p][v] = nq;
      for (i = 0; i <= 27; i++) T[nq][i] = T[q][i];
    }
  }
}
vector<int> g[400010], gg[400010];
int in[400010], out[400010], dfn, F[400010][30], h[400010];
long long st[400010], siz1[400010], siz2[400010];
int c, C[N * 2];
void dfs(int u, int f) {
  int i, v;
  h[u] = h[f] + 1;
  F[u][0] = f;
  for (i = 1; i <= 23; i++) F[u][i] = F[F[u][i - 1]][i - 1];
  in[u] = ++dfn;
  for (i = 0; i < g[u].size(); i++) {
    v = g[u][i];
    if (v != f) dfs(v, u);
  }
  out[u] = dfn;
}
int lca(int a, int b) {
  if (h[a] < h[b]) swap(a, b);
  int i;
  for (i = 22; i >= 0; i--)
    if (h[F[a][i]] >= h[b]) a = F[a][i];
  if (a == b) return a;
  for (i = 22; i >= 0; i--) {
    if (F[a][i] != F[b][i]) {
      a = F[a][i];
      b = F[b][i];
    }
  }
  return F[a][0];
}
int cmp(int a, int b) { return in[a] < in[b]; }
void build() {
  int i, j, m = c, top = 0;
  sort(C + 1, C + 1 + c, cmp);
  for (i = 1; i < m; i++) C[++c] = lca(C[i], C[i + 1]);
  sort(C + 1, C + 1 + c, cmp);
  c = unique(C + 1, C + 1 + c) - C - 1;
  for (i = 1; i <= c; i++) {
    if (top == 0)
      st[++top] = C[i];
    else {
      while (out[st[top]] < in[C[i]]) top--;
      gg[st[top]].push_back(C[i]);
      st[++top] = C[i];
    }
  }
}
void clear() {
  int i, a;
  for (i = 1; i <= c; i++) {
    a = C[i];
    siz1[a] = siz2[a] = 0;
    gg[a].clear();
  }
}
long long ans;
void solve(int u) {
  int i, v;
  for (i = 0; i < gg[u].size(); i++) {
    v = gg[u][i];
    solve(v);
    siz1[u] += siz1[v];
    siz2[u] += siz2[v];
    ans -= siz1[v] * siz2[v] * len[u];
  }
  ans += siz1[u] * siz2[u] * len[u];
}
int main() {
  int i, n, q, x, a, b;
  scanf("%d%d", &n, &q);
  scanf("%s", s + 1);
  last = cnt = 1;
  for (i = n; i; i--) build(s[i] - 'a');
  int p = 1;
  for (i = n; i; i--) {
    p = T[p][s[i] - 'a'];
    pos[i] = p;
  }
  for (i = 2; i <= cnt; i++) g[fa[i]].push_back(i);
  dfs(1, 0);
  while (q--) {
    scanf("%d%d", &a, &b);
    c = 0;
    for (i = 1; i <= a; i++) {
      scanf("%d", &x);
      x = pos[x];
      C[++c] = x;
      siz1[x]++;
    }
    for (i = 1; i <= b; i++) {
      scanf("%d", &x);
      x = pos[x];
      C[++c] = x;
      siz2[x]++;
    }
    build();
    ans = 0;
    solve(C[1]);
    cout << ans << endl;
    clear();
  }
  return 0;
}