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 = 2e5 + 5;
char s[N];
int tp[N], rak[N], sa[N], tax[N], h[N], n, m, f[N][20], lg2[N], a[N], b[N],
    c[N << 1], sum[N << 1];
void radixSort() {
  for (int i = 1; i <= m; i++) tax[i] = 0;
  for (int i = 1; i <= n; i++) tax[rak[i]]++;
  for (int i = 1; i <= m; i++) tax[i] += tax[i - 1];
  for (int i = n; i; i--) sa[tax[rak[tp[i]]]--] = tp[i];
}
void build_sa() {
  n = strlen(s + 1);
  m = 125;
  for (int i = 1; i <= n; i++) rak[i] = s[i], tp[i] = i;
  radixSort();
  for (int p = 0, w = 1; p < n; w <<= 1, m = p) {
    p = 0;
    for (int i = 1; i <= w; i++) tp[++p] = n - w + i;
    for (int i = 1; i <= n; i++)
      if (sa[i] > w) tp[++p] = sa[i] - w;
    radixSort();
    swap(tp, rak);
    rak[sa[1]] = p = 1;
    for (int i = 2; i <= n; i++)
      rak[sa[i]] =
          (tp[sa[i]] == tp[sa[i - 1]] && tp[sa[i] + w] == tp[sa[i - 1] + w])
              ? p
              : ++p;
  }
  for (int i = 1, k = 0, j; i <= n; i++) {
    if (k) k--;
    j = sa[rak[i] - 1];
    while (s[i + k] == s[j + k]) k++;
    h[rak[i]] = k;
  }
}
void make_st() {
  for (int i = 1; i <= n; i++) f[i][0] = h[i];
  int range = lg2[n];
  for (int i = 1; i <= range; i++)
    for (int j = 1; j + (1 << i) - 1 <= n; j++)
      f[j][i] = min(f[j][i - 1], f[j + (1 << (i - 1))][i - 1]);
}
int get_lcp(int l, int r) {
  if (l == r) return n - sa[l] + 1;
  if (l > r) swap(l, r);
  l++;
  int range = lg2[r - l + 1];
  return min(f[l][range], f[r - (1 << range) + 1][range]);
}
struct node {
  int id, l;
} st[N];
int main() {
  for (int i = 2; i < N; i++)
    lg2[i] = lg2[i - 1] + (i == (1 << (lg2[i - 1] + 1)));
  int q;
  scanf("%d %d %s", &n, &q, s + 1);
  build_sa();
  make_st();
  while (q--) {
    int al, bl;
    scanf("%d %d", &al, &bl);
    for (int i = 1, x; i <= al; i++) {
      scanf("%d", &x);
      a[i] = rak[x];
    }
    for (int i = 1, x; i <= bl; i++) {
      scanf("%d", &x);
      b[i] = rak[x];
    }
    sort(a + 1, a + 1 + al);
    sort(b + 1, b + 1 + bl);
    long long ans = 0;
    int top = 0, cl = al + bl;
    sum[0] = 0;
    for (int i = 1, j = 1, k = 1; i <= cl; i++) {
      if ((j <= al && a[j] <= b[k]) || k > bl)
        c[i] = a[j++], sum[i] = sum[i - 1];
      else
        c[i] = b[k++], sum[i] = sum[i - 1] + 1;
    }
    for (int i = 1; i <= cl; i++) {
      int l = i;
      while (top && get_lcp(c[i], c[i - 1]) <
                        get_lcp(c[st[top].id], c[st[top].id - 1])) {
        l = st[top].l;
        ans += (1ll * (sum[st[top].id - 1] - sum[max(st[top].l - 2, 0)]) *
                    ((i - 1 - st[top].id + 1) -
                     (sum[i - 1] - sum[st[top].id - 1])) +
                (1ll *
                 (st[top].id - 1 - st[top].l + 1 + 1 -
                  (sum[st[top].id - 1] - sum[max(st[top].l - 2, 0)])) *
                 (sum[i - 1] - sum[st[top].id - 1]))) *
               get_lcp(c[st[top].id], c[st[top].id - 1]);
        top--;
      }
      st[++top] = {i, l};
    }
    while (top) {
      ans += (1ll * (sum[st[top].id - 1] - sum[max(st[top].l - 2, 0)]) *
                  ((cl - st[top].id + 1) - (sum[cl] - sum[st[top].id - 1])) +
              (1ll *
               (st[top].id - 1 - st[top].l + 1 + 1 -
                (sum[st[top].id - 1] - sum[max(st[top].l - 2, 0)])) *
               (sum[cl] - sum[st[top].id - 1]))) *
             get_lcp(c[st[top].id], c[st[top].id - 1]);
      top--;
    }
    printf("%lld\n", ans);
  }
  return 0;
}