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 maxn = 4e5 + 10;
int n, q;
char A[maxn];
int m, sa[maxn], rk[maxn], ht[maxn], tp[maxn], bf[maxn], buk[maxn], lg[maxn],
    st[18][maxn];
void radix_sort() {
  memset(buk, 0, (m + 1) << 2);
  for (int i = (1), iend = (n); i <= iend; ++i) buk[bf[i] = rk[tp[i]]]++;
  for (int i = (1), iend = (m); i <= iend; ++i) buk[i] += buk[i - 1];
  for (int i = (n), iend = (1); i >= iend; --i) sa[buk[bf[i]]--] = tp[i];
}
void suffix_sort() {
  m = 150;
  for (int i = (1), iend = (n); i <= iend; ++i) rk[i] = A[i], tp[i] = i;
  radix_sort();
  for (int k = 1; k <= n; k <<= 1) {
    int num = 0;
    for (int i = (n), iend = (n - k + 1); i >= iend; --i) tp[++num] = i;
    for (int i = (1), iend = (n); i <= iend; ++i)
      if (sa[i] > k) tp[++num] = sa[i] - k;
    radix_sort();
    swap(rk, tp), rk[sa[1]] = 1, num = 1;
    for (int i = (2), iend = (n); i <= iend; ++i)
      rk[sa[i]] = num +=
          tp[sa[i]] != tp[sa[i - 1]] || tp[sa[i] + k] != tp[sa[i - 1] + k];
    if (n == num) break;
    m = num;
  }
  for (int i = 1, j = 0; i <= n; i++) {
    if (j) j--;
    while (A[i + j] == A[sa[rk[i] - 1] + j]) j++;
    ht[rk[i]] = j;
  }
  for (int i = (2), iend = (n); i <= iend; ++i) lg[i] = lg[i >> 1] + 1;
  for (int i = (1), iend = (n); i <= iend; ++i) st[0][i] = ht[i];
  for (int i = (1), iend = (17); i <= iend; ++i)
    for (int j = (1), jend = (n - (1 << i) + 1); j <= jend; ++j)
      st[i][j] = min(st[i - 1][j], st[i - 1][j + (1 << (i - 1))]);
}
int qmin(int l, int r) {
  int k = lg[r - l + 1];
  return min(st[k][l], st[k][r - (1 << k) + 1]);
}
int main() {
  scanf("%d%d%s", &n, &q, A + 1);
  suffix_sort();
  while (q--) {
    static int st[maxn], lef[maxn], rig[maxn], X[maxn], Y[maxn], V[maxn],
        val[maxn];
    int top = 0;
    int cx, cy;
    scanf("%d%d", &cx, &cy);
    int sz = cx + cy;
    for (int i = (1), iend = (cx); i <= iend; ++i)
      scanf("%d", &X[i]), V[i] = X[i] = rk[X[i]];
    for (int i = (1), iend = (cy); i <= iend; ++i)
      scanf("%d", &Y[i]), V[cx + i] = Y[i] = rk[Y[i]];
    sort(X + 1, X + cx + 1), sort(Y + 1, Y + cy + 1), sort(V + 1, V + sz + 1),
        sz = unique(V + 1, V + sz + 1) - V - 1;
    long long ans = 0;
    set<int> vis;
    for (int i = (1), iend = (cx); i <= iend; ++i) vis.insert(X[i]);
    for (int i = (1), iend = (cy); i <= iend; ++i)
      if (vis.count(Y[i])) ans += n - sa[Y[i]] + 1;
    set<int> visy;
    for (int i = (1), iend = (cy); i <= iend; ++i) visy.insert(Y[i]);
    for (int i = (2), iend = (sz); i <= iend; ++i)
      val[i] = qmin(V[i - 1] + 1, V[i]);
    st[0] = 1;
    for (int i = (2), iend = (sz); i <= iend; ++i) {
      while (top && val[st[top]] > val[i]) top--;
      lef[i] = st[top] + 1, st[++top] = i;
    }
    st[top = 0] = sz + 1;
    for (int i = (sz), iend = (2); i >= iend; --i) {
      while (top && val[st[top]] >= val[i]) top--;
      rig[i] = st[top] - 1, st[++top] = i;
    }
    for (int i = (2), iend = (sz); i <= iend; ++i) {
      long long c1 = 1ll *
                     (upper_bound(X + 1, X + cx + 1, V[i - 1]) -
                      lower_bound(X + 1, X + cx + 1, V[lef[i] - 1])) *
                     (upper_bound(Y + 1, Y + cy + 1, V[rig[i]]) -
                      lower_bound(Y + 1, Y + cy + 1, V[i]));
      long long c2 = 1ll *
                     (upper_bound(X + 1, X + cx + 1, V[rig[i]]) -
                      lower_bound(X + 1, X + cx + 1, V[i])) *
                     (upper_bound(Y + 1, Y + cy + 1, V[i - 1]) -
                      lower_bound(Y + 1, Y + cy + 1, V[lef[i] - 1]));
      ans += 1ll * val[i] * (c1 + c2);
    }
    printf("%lld\n", ans);
  }
  return 0;
}