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 LG[200010];
long long ans = 0;
struct node {
  int x, y;
  bool operator<(const node zz) const { return x < zz.x; }
} e[200010 << 1];
struct cc {
  int x, y;
  long long z;
} st[200010], ls[200010];
struct SA {
  char s[200010];
  int sa[200010], t[200010], t2[200010], c[200010], rak[200010], height[200010],
      f[200010][20], num[200010 << 1], n, m;
  void build_sa() {
    int i, *x = t, *y = t2;
    for (i = 0; i < m; i++) c[i] = 0;
    for (i = 0; i < n; i++) c[x[i] = s[i]]++;
    for (i = 1; i < m; i++) c[i] += c[i - 1];
    for (i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;
    for (int k = 1; k <= n; k <<= 1) {
      int p = 0;
      for (i = n - k; i < n; i++) y[p++] = i;
      for (i = 0; i < n; i++)
        if (sa[i] >= k) y[p++] = sa[i] - k;
      for (i = 0; i < m; i++) c[i] = 0;
      for (i = 0; i < n; i++) c[x[y[i]]]++;
      for (i = 1; i < m; i++) c[i] += c[i - 1];
      for (i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
      swap(x, y);
      p = 1;
      x[sa[0]] = 0;
      for (i = 1; i < n; i++)
        x[sa[i]] = y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k]
                       ? p - 1
                       : p++;
      if (p >= n) break;
      m = p;
    }
  }
  void getheight() {
    int i, j, k = 0;
    for (i = 0; i < n; i++) rak[sa[i]] = i;
    for (i = 0; i < n; i++) {
      if (k) k--;
      if (!rak[i]) continue;
      j = sa[rak[i] - 1];
      while (s[i + k] == s[j + k]) k++;
      height[rak[i]] = k;
    }
    for (int i = 0; i < n; i++) f[i][0] = height[i];
    for (int k = 1; (1 << k) < n; k++)
      for (int i = 0; i < n; i++)
        if (i + (1 << k - 1) < n)
          f[i][k] = min(f[i][k - 1], f[i + (1 << k - 1)][k - 1]);
  }
  int LCP(int l, int r) {
    if (l > r) swap(l, r);
    if (l == r) return n - sa[r] - 1;
    l++;
    int k = LG[r - l];
    return min(f[l][k], f[r - (1 << k) + 1][k]);
  }
  void work(int len) {
    int cnt = 0;
    for (int i = 1, t = 0; i <= len; i++)
      if (!e[i].y)
        num[i] = ++cnt, t = i;
      else
        num[i] = t;
    st[0] = {-1, 0, 0};
    int k = 0;
    for (int i = 1; i <= len; i++)
      if (e[i].y) {
        int j = i - 1, tot = 0;
        for (; j > 0 && !e[j].y; j--)
          ls[++tot] = cc{LCP(e[j].x, e[i].x), 1, LCP(e[j].x, e[i].x)};
        if (num[j]) {
          int t = LCP(e[j].x, e[i].x);
          while (st[k].x > t) ls[++tot] = {t, st[k].y, 1ll * t * st[k].y}, k--;
        }
        for (int i = tot; i; i--) {
          st[++k] = ls[i];
          if (st[k - 1].x == st[k].x) {
            k--;
            st[k].y += st[k + 1].y;
            st[k].z += st[k + 1].z;
          } else
            st[k].z += st[k - 1].z;
        }
        ans += st[k].z;
      }
  }
  void sol() {
    int k, l;
    scanf("%d%d", &k, &l);
    for (int i = 1, x; i <= k; i++) {
      scanf("%d", &x);
      e[i] = {rak[x - 1], 0};
    }
    for (int i = 1, x; i <= l; i++) {
      scanf("%d", &x);
      e[i + k] = {rak[x - 1], 1};
    }
    ans = 0;
    sort(e + 1, e + k + l + 1);
    work(k + l);
    reverse(e + 1, e + k + l + 1);
    work(l + k);
    printf("%lld\n", ans);
  }
} S;
int main(int argc, char const *argv[]) {
  for (int i = 1; (1 << i) < 200010; i++) LG[1 << i] = 1;
  for (int i = 1; i < 200010; i++) LG[i] += LG[i - 1];
  int n, q;
  scanf("%d%d", &n, &q);
  scanf("%s", S.s);
  S.s[n] = '$';
  n++;
  S.n = n;
  S.m = 300;
  S.build_sa();
  S.getheight();
  while (q--) S.sol();
  return 0;
}