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>
#pragma GCC optimize("Ofast,inline,unroll-loops,fast-math")
using namespace std;
class Node {
 public:
  int s[26];
  int f;
  int l;
  Node(void) : f(-1), l(0) { return; }
};
char a[400020];
Node f[400020];
int c, p[400020];
vector<int> e[400020], r[400020], h;
int l[400020], d[400020], b[19][400020];
int g[2][400020];
inline int ID(char c) { return c - 97; }
bool Compare(int x, int y) { return l[x] < l[y]; }
void InitSAM(void) {
  c = 1;
  return;
}
void InsertSAM(int i, int k) {
  static int d;
  int p, q, b;
  p = d;
  f[d = c++].l = f[p].l + 1;
  ::p[i] = d;
  for (; p != -1 && !f[p].s[k]; p = f[p].f) f[p].s[k] = d;
  if (p == -1)
    f[d].f = 0;
  else {
    q = f[p].s[k];
    if (f[q].l == f[p].l + 1)
      f[d].f = q;
    else {
      f[b = c++] = f[q];
      f[b].l = f[p].l + 1;
      f[d].f = f[q].f = b;
      for (; p != -1 && f[p].s[k] == q; p = f[p].f) f[p].s[k] = b;
    }
  }
  return;
}
void DFS(int x) {
  static int k;
  int i;
  l[x] = k++;
  for (i = 1; i < 19; i++) b[i][x] = b[i - 1][b[i - 1][x]];
  for (i = 0; i < (signed)e[x].size(); i++) {
    d[e[x][i]] = d[x] + 1;
    b[0][e[x][i]] = x;
    DFS(e[x][i]);
  }
  return;
}
int GetLCA(int u, int v) {
  int i, t;
  if (d[u] < d[v]) swap(u, v);
  t = d[u] - d[v];
  for (i = 0; i < 19; i++)
    if (t & (1 << i)) u = b[i][u];
  if (u == v) return u;
  for (i = 19 - 1; i > -1; i--)
    if (b[i][u] != b[i][v]) {
      u = b[i][u];
      v = b[i][v];
    }
  return b[0][u];
}
long long Solve(int x) {
  int i, p, q;
  long long o;
  for (i = o = p = q = 0; i < (signed)r[x].size(); i++) {
    o += Solve(r[x][i]);
    p += g[0][r[x][i]];
    q += g[1][r[x][i]];
  }
  o += (long long)f[x].l * g[0][x] * q;
  o += (long long)f[x].l * g[1][x] * p;
  o += (long long)f[x].l * g[0][x] * g[1][x];
  for (i = 0; i < (signed)r[x].size(); i++) {
    o += (long long)f[x].l * g[0][r[x][i]] * (q - g[1][r[x][i]]);
  }
  g[0][x] += p;
  g[1][x] += q;
  return o;
}
int main(void) {
  int n, m, t, x, y;
  int i;
  scanf("%d %d %s", &n, &m, a);
  InitSAM();
  for (i = n - 1; i > -1; i--) InsertSAM(n - i - 1, ID(a[i]));
  for (i = 1; i < c; i++) e[f[i].f].push_back(i);
  DFS(0);
  while (m--) {
    scanf("%d %d", &x, &y);
    h.clear();
    for (i = 0; i < x; i++) {
      scanf("%d", &t);
      h.push_back(p[n - t]);
      g[0][p[n - t]]++;
    }
    for (i = 0; i < y; i++) {
      scanf("%d", &t);
      h.push_back(p[n - t]);
      g[1][p[n - t]]++;
    }
    sort(h.begin(), h.end(), Compare);
    for (i = 1; i < x + y; i++) h.push_back(GetLCA(h[i], h[i - 1]));
    sort(h.begin(), h.end(), Compare);
    h.erase(unique(h.begin(), h.end()), h.end());
    for (i = 1; i < (signed)h.size(); i++)
      r[GetLCA(h[i], h[i - 1])].push_back(h[i]);
    printf("%lld\n", Solve(h[0]));
    for (i = 0; i < (signed)h.size(); i++) {
      r[h[i]].clear();
      g[0][h[i]] = g[1][h[i]] = 0;
    }
  }
  return 0;
}