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;
using ll = long long;
inline int read();
const int M = 200016, MOD = 1000000007;
namespace SA {
void st_init(int *arr, int n);
int sa[M], rk[M], height[M];
int t1[M], t2[M], c[M];
void build(char *str, int n, int m) {
  str[n] = 0;
  n++;
  int i, j, p, *x = t1, *y = t2;
  for (i = 0; i < m; i++) c[i] = 0;
  for (i = 0; i < n; i++) c[x[i] = str[i]]++;
  for (i = 1; i < m; i++) c[i] += c[i - 1];
  for (i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;
  for (j = 1; j <= n; j <<= 1) {
    p = 0;
    for (i = n - j; i < n; i++) y[p++] = i;
    for (i = 0; i < n; i++)
      if (sa[i] >= j) y[p++] = sa[i] - j;
    for (i = 0; i < m; i++) c[i] = 0;
    for (i = 0; i < n; i++) c[x[y[i]]]++;
    for (i = 1; i < m; i++) c[i] += c[i - 1];
    for (i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
    swap(x, y);
    p = 1;
    x[sa[0]] = 0;
    for (i = 1; i < n; i++)
      x[sa[i]] = (y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + j] == y[sa[i] + j])
                     ? p - 1
                     : p++;
    if (p >= n) break;
    m = p;
  }
  n--;
  for (int i = 0; i <= n; i++) rk[sa[i]] = i;
  for (int i = 0, j = 0, k = 0; i < n; i++) {
    if (k) k--;
    j = sa[rk[i] - 1];
    while (str[i + k] == str[j + k]) k++;
    height[rk[i]] = k;
  }
  st_init(height, n);
}
int lg[M], _n;
int table[20][M];
void st_init(int *arr, int n) {
  _n = n;
  if (!lg[0]) {
    lg[0] = -1;
    for (int i = 1; i < M; i++) lg[i] = lg[i / 2] + 1;
  }
  for (int i = 1; i <= n; ++i) table[0][i] = arr[i];
  for (int i = 1; i <= lg[n]; ++i)
    for (int j = 1; j <= n; ++j)
      if (j + (1 << i) - 1 <= n)
        table[i][j] = min(table[i - 1][j], table[i - 1][j + (1 << (i - 1))]);
}
int mih(int l, int r) {
  int t = lg[r - l + 1];
  return min(table[t][l], table[t][r - (1 << t) + 1]);
}
};  // namespace SA
struct MonoStack {
  ll sum = 0, r = 0;
  pair<ll, ll> save[M];
  void init() { sum = r = 0; }
  ll solve(ll key, ll add) {
    while (r && key <= save[r - 1].first) {
      add += save[--r].second;
      sum -= save[r].first * save[r].second;
    }
    if (add) {
      sum += key * add;
      save[r++] = {key, add};
    }
    return sum;
  }
} st[2];
char str[M];
int A[M], B[M];
int main(void) {
  int n = read(), q = read();
  scanf("%s", str);
  SA::build(str, n, 128);
  using SA::height;
  using SA::mih;
  using SA::rk;
  while (q--) {
    st[0].init(), st[1].init();
    int la = read(), lb = read();
    for (int i = 1; i <= la; ++i) A[i] = read() - 1;
    for (int i = 1; i <= lb; ++i) B[i] = read() - 1;
    ll ans = 0;
    for (int i = 1, j = 1; i <= la && j <= lb;) {
      if (A[i] == B[j])
        ans += n - A[i], ++i, ++j;
      else if (A[i] < B[j])
        ++i;
      else
        ++j;
    }
    vector<pair<int, int>> ops;
    for (int i = 1; i <= la; ++i)
      ops.emplace_back(rk[A[i]] + 1, 0), ops.emplace_back(rk[A[i]], 2);
    for (int i = 1; i <= lb; ++i)
      ops.emplace_back(rk[B[i]] + 1, 1), ops.emplace_back(rk[B[i]], 3);
    sort(ops.begin(), ops.end());
    for (int i = 0, lst = ops[0].first; i < (int)ops.size(); ++i) {
      int ti = ops[i].first, op = ops[i].second;
      if (op < 2) {
        st[op].solve(height[ti], 1);
        st[op ^ 1].solve(height[ti], 0);
      } else {
        ans += st[3 - op].solve(mih(lst, ti), 0);
        st[op - 2].solve(mih(lst, ti), 0);
      }
      lst = ti;
    }
    cout << ans << "\n";
  }
  return 0;
}
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}