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 = 214514;
char s[N];
int sa[N], ct[N], rk[N], tp[N], h[N], lc[N][19];
int msk, kkr = 514, hkn, n, k, l, q, po;
long long as, tt[2], st[N], w[N][2];
pair<int, int> a[N << 1];
void getsa() {
  scanf("%d%d%s", &n, &q, s + 1);
  for (int i = (1); i <= (n); i++) ct[rk[i] = s[i]]++;
  for (int i = (1); i <= (kkr); i++) ct[i] += ct[i - 1];
  for (int i = (n); i >= (1); i--) sa[ct[rk[i]]--] = i;
  for (int k = 1;; k <<= 1) {
    msk = 0;
    for (int i = (0); i <= (kkr); i++) ct[i] = 0;
    for (int i = (n - k + 1); i <= (n); i++) tp[++msk] = i;
    for (int i = (1); i <= (n); i++)
      if (sa[i] > k) tp[++msk] = sa[i] - k;
    for (int i = (1); i <= (n); i++) ct[rk[i]]++;
    for (int i = (1); i <= (kkr); i++) ct[i] += ct[i - 1];
    for (int i = (n); i >= (1); i--) sa[ct[rk[tp[i]]]--] = tp[i];
    swap(tp, rk), msk = rk[sa[1]] = 1;
    for (int i = (2); i <= (n); i++)
      rk[sa[i]] =
          tp[sa[i]] == tp[sa[i - 1]] && tp[sa[i] + k] == tp[sa[i - 1] + k]
              ? msk
              : ++msk;
    kkr = msk;
    if (kkr >= n) break;
  }
  for (int i = 1, j = 0; i <= n; i++) {
    if (j) j--;
    while (s[i + j] == s[sa[rk[i] - 1] + j]) j++;
    lc[rk[i]][0] = j;
  }
}
int main() {
  getsa();
  int x;
  for (int j = (1); j <= (17); j++)
    for (int i = (1); i <= (n); i++)
      lc[i][j] = min(lc[i][j - 1], lc[i + (1 << (j - 1))][j - 1]);
  for (int ad = (1); ad <= (q); ad++) {
    tt[0] = tt[1] = po = as = 0;
    scanf("%d%d", &k, &l);
    for (int i = (1); i <= (k); i++) scanf("%d", &x), a[i] = {rk[x], 0};
    for (int i = (1); i <= (l); i++) scanf("%d", &x), a[i + k] = {rk[x], 1};
    sort(a + 1, a + k + l + 1);
    int ls = a[1].first;
    for (int i = (2); i <= (k + l); i++) {
      long long x, al[2] = {0, 0}, nw = a[i].first;
      al[a[i - 1].second] = 1;
      if (a[i].first == ls)
        x = n - sa[nw] + 1;
      else {
        for (int j = (17); j >= (0); j--)
          if ((1 << j) & (nw - ls)) {
            x = min(lc[ls + 1][j], lc[nw - (1 << j) + 1][j]);
            break;
          }
      }
      ls = nw;
      for (; po && x <= st[po]; po--)
        for (int j = (0); j <= (1); j++)
          tt[j] -= w[po][j] * st[po], al[j] += w[po][j];
      st[++po] = x;
      for (int j = (0); j <= (1); j++) w[po][j] = al[j], tt[j] += x * al[j];
      as += tt[a[i].second ^ 1];
    }
    printf("%lld\n", as);
  }
}