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 text[200009];
int lcp[200009];
int RA[200009], tempRA[200009];
int SA[200009], tempSA[200009];
int c[200009];
void countingSort(int n, int k) {
  int i, sum, maxi = max(260, n);
  memset(c, 0, (maxi + 5) * sizeof(int));
  for (i = 0; i < n; i++) c[i + k < n ? RA[i + k] : 0]++;
  for (i = sum = 0; i < maxi; i++) {
    int t = c[i];
    c[i] = sum;
    sum += t;
  }
  for (i = 0; i < n; i++)
    tempSA[c[SA[i] + k < n ? RA[SA[i] + k] : 0]++] = SA[i];
  for (i = 0; i < n; i++) SA[i] = tempSA[i];
}
void constructSA(int n) {
  int i, k, r;
  for (i = 0; i < n; i++) RA[i] = text[i] - 29;
  for (i = 0; i < n; i++) SA[i] = i;
  for (k = 1; k < n; k <<= 1) {
    countingSort(n, k);
    countingSort(n, 0);
    tempRA[SA[0]] = r = 0;
    for (i = 1; i < n; i++)
      tempRA[SA[i]] =
          (RA[SA[i]] == RA[SA[i - 1]] && RA[SA[i] + k] == RA[SA[i - 1] + k])
              ? r
              : ++r;
    for (i = 0; i < n; i++) RA[i] = tempRA[i];
  }
}
void computeLCP(int n) {
  for (int i = 0, k = 0; i < n; i++) {
    if (k > 0) k--;
    if (RA[i] == n - 1) {
      lcp[n - 1] = -1;
      k = 0;
      continue;
    }
    int j = SA[RA[i] + 1];
    while (text[i + k] == text[j + k]) k++;
    lcp[RA[i]] = k;
  }
}
int comp(int a, int b) { return min(a, b); }
int st[200009][20];
class SparseTable {
 private:
  int sz;

 public:
  SparseTable(int* ini, int* en) {
    sz = int(en - ini);
    for (int i = 0; i < sz; i++) st[i][0] = ini[i];
    for (int j = 1; 1 << j <= sz; j++)
      for (int i = 0; i + (1 << j) <= sz; i++)
        st[i][j] = comp(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
  }
  int query(int l, int r) {
    int k = (int)floor(log((double)r - l + 1) / log(2.0));
    return comp(st[l][k], st[r - (1 << k) + 1][k]);
  }
};
int main() {
  int n, q;
  scanf("%d %d %s", &n, &q, &text);
  text[n] = '$';
  constructSA(n + 1);
  computeLCP(n + 1);
  SparseTable table(lcp, lcp + n + 1);
  while (q--) {
    int a, b;
    scanf("%d %d", &a, &b);
    vector<int> va(a), vb(b);
    for (int i = 0; i < a; i++) {
      scanf("%d", &va[i]);
      va[i]--;
      va[i] = RA[va[i]];
    }
    for (int i = 0; i < b; i++) {
      scanf("%d", &vb[i]);
      vb[i]--;
      vb[i] = RA[vb[i]];
    }
    sort(va.begin(), va.end());
    sort(vb.begin(), vb.end());
    int ini = 0;
    long long ans = 0, acum = 0;
    vector<pair<int, int> > lst;
    for (int i = 0; i < a; i++) {
      if (ini > 0) {
        int k = table.query(vb[ini - 1], va[i] - 1);
        int t = 0;
        while (!lst.empty() && lst.back().first >= k) {
          acum -= lst.back().first * 1ll * lst.back().second;
          t += lst.back().second;
          lst.pop_back();
        }
        if (t > 0) {
          lst.push_back(pair<int, int>(k, t));
          acum += lst.back().first * 1ll * lst.back().second;
        }
      }
      while (ini < b && vb[ini] < va[i]) {
        int k = table.query(vb[ini], va[i] - 1);
        int t = 1;
        while (!lst.empty() && lst.back().first >= k) {
          acum -= lst.back().first * 1ll * lst.back().second;
          t += lst.back().second;
          lst.pop_back();
        }
        lst.push_back(pair<int, int>(k, t));
        acum += lst.back().first * 1ll * lst.back().second;
        ini++;
      }
      while (ini < b && vb[ini] == va[i]) {
        int k = n - SA[va[i]];
        int t = 1;
        while (!lst.empty() && lst.back().first >= k) {
          acum -= lst.back().first * 1ll * lst.back().second;
          t += lst.back().second;
          lst.pop_back();
        }
        lst.push_back(pair<int, int>(k, t));
        acum += lst.back().first * 1ll * lst.back().second;
        ini++;
      }
      ans += acum;
    }
    swap(va, vb);
    swap(a, b);
    lst.clear();
    acum = 0;
    ini = 0;
    for (int i = 0; i < a; i++) {
      if (ini > 0) {
        int k = table.query(vb[ini - 1], va[i] - 1);
        int t = 0;
        while (!lst.empty() && lst.back().first >= k) {
          acum -= lst.back().first * 1ll * lst.back().second;
          t += lst.back().second;
          lst.pop_back();
        }
        if (t > 0) {
          lst.push_back(pair<int, int>(k, t));
          acum += lst.back().first * 1ll * lst.back().second;
        }
      }
      while (ini < b && vb[ini] < va[i]) {
        int k = table.query(vb[ini], va[i] - 1);
        int t = 1;
        while (!lst.empty() && lst.back().first >= k) {
          acum -= lst.back().first * 1ll * lst.back().second;
          t += lst.back().second;
          lst.pop_back();
        }
        lst.push_back(pair<int, int>(k, t));
        acum += lst.back().first * 1ll * lst.back().second;
        ini++;
      }
      ans += acum;
      while (ini < b && vb[ini] == va[i]) {
        int k = n - SA[va[i]];
        int t = 1;
        while (!lst.empty() && lst.back().first >= k) {
          acum -= lst.back().first * 1ll * lst.back().second;
          t += lst.back().second;
          lst.pop_back();
        }
        lst.push_back(pair<int, int>(k, t));
        acum += lst.back().first * 1ll * lst.back().second;
        ini++;
      }
    }
    printf("%lld\n", ans);
    va.clear();
    vb.clear();
  }
}