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;
#pragma warning(disable : 4996)
const int N = 2e5 + 5, L = 19;
string s;
int cl[L][N], sf[L][N], ct[N], pos[N], lcp[N], t[L][N], g[N];
pair<int, int> a[N + N];
int *suf = sf[L - 1];
int que(int l, int r) {
  int k = g[r - l + 1];
  return min(t[k][l], t[k][r - (1 << k) + 1]);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  for (int i = 2; i < N; i++) g[i] = g[i >> 1] + 1;
  int n, q;
  cin >> n >> q >> s;
  s += "#";
  n++;
  for (int i = 0; i < n; i++) {
    ct[s[i]]++;
  }
  for (int i = 0; i < N - 1; i++) {
    ct[i + 1] += ct[i];
  }
  for (int i = n - 1; i >= 0; i--) {
    sf[0][--ct[s[i]]] = i;
    cl[0][i] = s[i];
  }
  for (int i = 1; i < L; i++) {
    for (int j = 0; j < n; j++) {
      sf[i - 1][j] = (sf[i - 1][j] - (1 << (i - 1)) % n + n) % n;
    }
    fill(ct, ct + N, 0);
    for (int j = 0; j < n; j++) {
      ct[cl[i - 1][sf[i - 1][j]]]++;
    }
    for (int j = 0; j < N - 1; j++) {
      ct[j + 1] += ct[j];
    }
    for (int j = n - 1; j >= 0; j--) {
      sf[i][--ct[cl[i - 1][sf[i - 1][j]]]] = sf[i - 1][j];
    }
    for (int j = 0, k = 0; j < n; j++) {
      cl[i][sf[i][j]] = k;
      if (j + 1 < n && (cl[i - 1][sf[i][j]] != cl[i - 1][sf[i][j + 1]] ||
                        cl[i - 1][(sf[i][j] + (1 << (i - 1))) % n] !=
                            cl[i - 1][(sf[i][j + 1] + (1 << (i - 1))) % n])) {
        k++;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    pos[suf[i]] = i;
  }
  for (int i = 0, j = 0; i < n; i++) {
    int p = pos[i];
    if (p == n - 1) {
      j = 0;
      continue;
    }
    int q = suf[p + 1];
    while (i + j < n && q + j < n && s[i + j] == s[q + j]) {
      j++;
    }
    lcp[p] = j;
    j = max(0, j - 1);
  }
  for (int i = 0; i < n; i++) {
    t[0][i] = lcp[i];
  }
  for (int i = 1; i < L; i++) {
    for (int j = 0; j + (1 << i) <= n; j++) {
      t[i][j] = min(t[i - 1][j], t[i - 1][j + (1 << (i - 1))]);
    }
  }
  for (int i = 0; i < q; i++) {
    int la, lb, ls;
    cin >> la >> lb;
    ls = 0;
    for (int j = 0; j < la; j++) {
      cin >> a[ls].first;
      a[ls].first--;
      a[ls++].second = 1;
    }
    for (int j = 0; j < lb; j++) {
      cin >> a[ls].first;
      a[ls].first--;
      a[ls++].second = 0;
    }
    sort(a, a + ls, [](pair<int, int> &x, pair<int, int> &y) {
      if (pos[x.first] == pos[y.first]) return x.second < y.second;
      return pos[x.first] < pos[y.first];
    });
    long long ans = 0;
    for (int j = 0; j < ls - 1; j++) {
      if (a[j].first == a[j + 1].first) ans += n - 1 - a[j].first;
    }
    vector<pair<int, int>> v;
    v.push_back({-1, 0});
    long long sum = 0;
    for (int j = 1; j < ls; j++) {
      if (a[j - 1].first != a[j].first) {
        int h = que(pos[a[j - 1].first], pos[a[j].first] - 1),
            x = a[j - 1].second;
        while (v.back().first >= h) {
          x += v.back().second;
          sum -= v.back().first * (long long)v.back().second;
          v.pop_back();
        }
        v.push_back({h, x});
        sum += x * (long long)h;
      }
      if (!a[j].second) ans += sum;
    }
    for (int j = 0; j < ls - 1; j++) {
      if (a[j].first == a[j + 1].first) swap(a[j], a[j + 1]);
    }
    v.clear();
    v.push_back({-1, 0});
    sum = 0;
    for (int j = ls - 2; j >= 0; j--) {
      if (a[j + 1].first != a[j].first) {
        int h = que(pos[a[j].first], pos[a[j + 1].first] - 1),
            x = a[j + 1].second;
        while (v.back().first >= h) {
          x += v.back().second;
          sum -= v.back().first * (long long)v.back().second;
          v.pop_back();
        }
        v.push_back({h, x});
        sum += x * (long long)h;
      }
      if (!a[j].second) ans += sum;
    }
    cout << ans << "\n";
  }
  return 0;
}