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 = 200050;
const int M = 2 * N;
const int L = 19;
char s[N];
pair<pair<int, int>, int> tmp[N];
int id[N * 2], sa[N], lcp[N], tot, rmq[N][L], logs[N];
void BuildSuffixArray() {
  s[++tot] = 'a' + 26;
  for (int i = 1; i <= tot; i++) id[i] = s[i] - 'a';
  for (int j = 0; j < 20; j++) {
    for (int i = 1; i <= tot; i++)
      tmp[i] = make_pair(make_pair(id[i], id[i + (1 << j)]), i);
    sort(tmp + 1, tmp + 1 + tot);
    int c = 0;
    for (int i = 1; i <= tot; i++) {
      id[tmp[i].second] = c + 1;
      c += tmp[i].first != tmp[i + 1].first;
    }
    if (c == tot) break;
  }
  for (int i = 1; i <= tot; i++) sa[id[i]] = i;
}
void BuildLCPArray() {
  int k = 0;
  for (int i = 1; i <= tot; i++) {
    if (id[i] == tot) continue;
    int j = sa[id[i] + 1];
    while (i + k <= tot && j + k <= tot && s[i + k] == s[j + k]) k++;
    lcp[id[i]] = k;
    if (k) k--;
  }
}
void BuildSparseTable() {
  for (int i = 2; i <= tot; i++) logs[i] = logs[i >> 1] + 1;
  for (int i = 1; i < tot; i++) rmq[i][0] = lcp[i];
  for (int j = 1; j < L; j++) {
    for (int i = 1; i < tot - (1 << j) + 1; i++) {
      rmq[i][j] = min(rmq[i][j - 1], rmq[i + (1 << (j - 1))][j - 1]);
    }
  }
}
int LCP(int l, int r) {
  l = id[l];
  r = id[r];
  if (l == r) return tot - sa[l];
  if (l > r) swap(l, r);
  r--;
  int k = logs[r - l + 1];
  return min(rmq[l][k], rmq[r - (1 << k) + 1][k]);
}
bool comp(pair<int, int> i, pair<int, int> j) {
  return id[i.first] < id[j.first];
}
int S[M], A[M], B[M];
int main() {
  int n, q, i, x, y;
  scanf("%i %i", &n, &q);
  scanf("%s", s + 1);
  tot = n;
  BuildSuffixArray();
  BuildLCPArray();
  BuildSparseTable();
  while (q--) {
    scanf("%i %i", &x, &y);
    vector<pair<int, int> > a;
    while (x--) scanf("%i", &i), a.push_back(make_pair(i, 0));
    while (y--) scanf("%i", &i), a.push_back(make_pair(i, 1));
    sort(a.begin(), a.end(), comp);
    long long suma = 0, sumb = 0, ans = 0;
    int c = 0;
    for (i = 1; i < a.size(); i++) {
      int tmp = LCP(a[i].first, a[i - 1].first);
      int asz = 0, bsz = 0;
      while (c && S[c] >= tmp) {
        suma -= (long long)A[c] * S[c];
        sumb -= (long long)B[c] * S[c];
        asz += A[c];
        bsz += B[c];
        c--;
      }
      suma += (long long)asz * tmp;
      sumb += (long long)bsz * tmp;
      S[++c] = tmp;
      A[c] = asz;
      B[c] = bsz;
      if (a[i - 1].second) {
        A[c]++;
        suma += tmp;
      } else {
        B[c]++;
        sumb += tmp;
      }
      if (a[i].second) {
        ans += sumb;
      } else {
        ans += suma;
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}