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 = 200010;
int x[N], y[N], c[N], h[N], m = 127, rk[N], sa[N], len;
void sort() {
  fill(c + 1, c + m + 1, 0);
  for (int i = 1; i <= len; i++) c[x[i]]++;
  for (int i = 1; i <= m; i++) c[i] += c[i - 1];
  for (int i = len; i >= 1; i--) sa[c[x[y[i]]]--] = y[i];
}
void SA(char s[]) {
  for (int i = 1; i <= len; i++) x[i] = s[i], y[i] = i;
  sort();
  for (int k = 1, num = 0; k <= len; k <<= 1, num = 0) {
    for (int i = len; i >= len - k + 1; i--) y[++num] = i;
    for (int i = 1; i <= len; i++)
      if (sa[i] > k) y[++num] = sa[i] - k;
    sort(), memcpy(y, x, sizeof(x)), num = x[sa[1]] = 1;
    for (int i = 2; i <= len; i++)
      x[sa[i]] = (y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k])
                     ? num
                     : ++num;
    if (num == len) break;
    m = num;
  }
}
int st[18][N], st1[18][N], lg[N], stk[N], L[N], R[N];
int Min(int a, int b) { return h[a] < h[b] ? a : b; }
void get(char s[]) {
  for (int i = 1; i <= len; i++) rk[sa[i]] = i;
  for (int i = 2; i <= len; i++) lg[i] = lg[i >> 1] + 1;
  for (int i = 1, k = 0; i <= len; i++)
    if (rk[i] != 1) {
      if (k) k--;
      while (s[sa[rk[i] - 1] + k] == s[i + k]) k++;
      h[rk[i]] = k, st[0][rk[i]] = st1[0][rk[i]] = rk[i];
    }
  for (int i = 1; i < 18; i++)
    for (int j = 1; j + (1 << i) - 1 <= len; j++)
      st[i][j] = Min(st[i - 1][j], st[i - 1][j + (1 << i - 1)]),
      st1[i][j] = Min(st1[i - 1][j + (1 << i - 1)], st1[i - 1][j]);
  int top = 1;
  stk[top] = 2;
  for (int i = 3; i <= len; i++) {
    while (top && h[stk[top]] > h[i]) R[stk[top--]] = i - 1;
    stk[++top] = i;
  }
  while (top) R[stk[top--]] = len;
  stk[top = 1] = len;
  for (int i = len - 1; i >= 2; i--) {
    while (top && h[stk[top]] >= h[i]) L[stk[top--]] = i + 1;
    stk[++top] = i;
  }
  while (top) L[stk[top--]] = 2;
}
int get(int l, int r) {
  int len = lg[r - l + 1];
  return Min(st[len][l], st[len][r - (1 << len) + 1]);
}
int get1(int l, int r) {
  int len = lg[r - l + 1];
  return Min(st1[len][r - (1 << len) + 1], st1[len][l]);
}
int T[2][N];
void upd(int t, int x, int y) {
  for (; x <= len; x += x & -x) T[t][x] += y;
}
int que(int t, int x) {
  int res = 0;
  for (; x; x -= x & -x) res += T[t][x];
  return res;
}
int que(int t, int l, int r) { return que(t, r) - que(t, l - 1); }
char s[N];
int main() {
  int q;
  scanf("%d%d%s", &len, &q, s + 1), SA(s), get(s);
  while (q--) {
    int A, B;
    scanf("%d%d", &A, &B);
    vector<int> a, b, c;
    for (int i = 1, t; i <= A; i++)
      scanf("%d", &t), t = rk[t], a.push_back(t), c.push_back(t), upd(0, t, 1);
    for (int i = 1, t; i <= B; i++)
      scanf("%d", &t), t = rk[t], b.push_back(t), c.push_back(t), upd(1, t, 1);
    sort(c.begin(), c.end());
    long long res = 0;
    for (int i = 0; i + 1 < c.size(); i++) {
      int tl = c[i], tr = c[i + 1];
      if (tl == tr) {
        res += len - sa[tl] + 1;
        continue;
      }
      int p = get(tl + 1, tr), l = L[get1(tl + 1, tr)], r = R[p];
      res += (long long)que(0, l - 1, p - 1) * que(1, p, r) * h[p];
      res += (long long)que(1, l - 1, p - 1) * que(0, p, r) * h[p];
    }
    for (auto t : a) upd(0, t, -1);
    for (auto t : b) upd(1, t, -1);
    printf("%lld\n", res);
  }
}