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;
char s[410000];
pair<int, int> f[410000][20];
int n, q, k, l, a[410000], b[410000], sa[410000], c[410000], x[410000],
    y[410000], height[410000], ra[410000];
void build_sa(int m) {
  memset(y, 0, sizeof(y));
  for (int i = 0; i <= m; i++) c[i] = 0;
  for (int i = 1; i <= n; i++) c[x[i] = s[i]]++;
  for (int i = 1; i <= m; i++) c[i] += c[i - 1];
  for (int i = n; i; i--) sa[c[x[i]]--] = i;
  for (int k = 1; k <= n; k <<= 1) {
    int p = 0;
    for (int i = n - k + 1; i <= n; i++) y[++p] = i;
    for (int i = 1; i <= n; i++)
      if (sa[i] > k) y[++p] = sa[i] - k;
    for (int i = 0; i <= m; i++) c[i] = 0;
    for (int i = 1; i <= n; i++) c[x[y[i]]]++;
    for (int i = 1; i <= m; i++) c[i] += c[i - 1];
    for (int i = n; i; i--) sa[c[x[y[i]]]--] = y[i];
    for (int i = 1; i <= n; i++) y[i] = x[i];
    p = 1;
    x[sa[1]] = 1;
    for (int i = 2; i <= n; i++)
      if (y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k])
        x[sa[i]] = p;
      else
        x[sa[i]] = ++p;
    if (p >= n) break;
    m = p;
  }
}
void getheight() {
  int k = 0, j;
  memset(height, 0, sizeof(height));
  for (int i = 1; i <= n; i++) ra[sa[i]] = i;
  for (int i = 1; i <= n; i++) {
    if (k) k--;
    j = sa[ra[i] - 1];
    while (s[i + k] == s[j + k]) k++;
    height[ra[i]] = k;
  }
}
int minnum(int l, int r) {
  int k = 0;
  while ((1 << (k + 1)) <= r - l + 1) k++;
  return min(f[l][k], f[r - (1 << k) + 1][k]).second;
}
void st() {
  for (int i = 1; i <= n; i++) f[i][0] = pair<int, int>(height[i], i);
  for (int j = 1; (1 << j) <= n; j++)
    for (int i = 1; i + (1 << j) - 1 <= n; i++)
      f[i][j] = min(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
}
long long C(int n) { return 1ll * n * (n - 1) / 2; }
long long sum(int l, int r) {
  if (a[l] == a[r]) return C(r - l + 1) * (n - sa[a[l]] + 1);
  int k = minnum(a[l] + 1, a[r]);
  int p = lower_bound(a + l, a + r + 1, k) - a;
  return 1ll * height[k] * (p - l) * (r - p + 1) + sum(l, p - 1) + sum(p, r);
}
long long work(int l, int r) {
  sort(a + l, a + r + 1);
  return sum(l, r);
}
int main() {
  scanf("%d%d", &n, &q);
  scanf("%s", s + 1);
  build_sa(200);
  getheight();
  st();
  for (int i = 1; i <= q; i++) {
    scanf("%d %d", &k, &l);
    long long ans = 0;
    for (int j = 1; j <= k; j++) {
      scanf("%d", &a[j]);
      a[j] = ra[a[j]];
    }
    ans -= work(1, k);
    for (int j = k + 1; j <= k + l; j++) {
      scanf("%d", &a[j]);
      a[j] = ra[a[j]];
    }
    ans -= work(k + 1, k + l);
    ans += work(1, k + l);
    printf("%lld\n", ans);
  }
  return 0;
}