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 s[500005], s1[500005], s2[500005];
char p[500005];
long long rak[500005], tp[500005], sa[500005], t[500005], h[500005], dp[500005],
    st[30][500005], aux[500005], n, m = 180;
struct zero {
  long long x, y;
} lib[500005];
void sort() {
  for (long long i = 1; i <= m; i++) t[i] = 0;
  for (long long i = 1; i <= n; i++) t[rak[i]]++;
  for (long long i = 1; i <= m; i++) t[i] += t[i - 1];
  for (long long i = n; i >= 1; i--) sa[t[rak[tp[i]]]--] = tp[i];
}
void work() {
  m = 280;
  for (long long i = 1; i <= n; i++) rak[i] = s[i] + 1, tp[i] = i;
  sort();
  for (long long w = 1, p = 0; p < n; m = p, w <<= 1) {
    p = 0;
    for (long long i = 1; i <= w; i++) tp[++p] = n - w + i;
    for (long long i = 1; i <= n; i++)
      if (sa[i] > w) tp[++p] = sa[i] - w;
    sort();
    for (long long i = 1; i <= n; i++) tp[i] = rak[i];
    p = 0;
    for (long long i = 1; i <= n; i++)
      rak[sa[i]] =
          (tp[sa[i - 1]] == tp[sa[i]] && tp[sa[i - 1] + w] == tp[sa[i] + w]
               ? p
               : ++p);
  }
  long long k = 0;
  for (long long i = 1; i <= n; i++) {
    if (k) k--;
    long long j = sa[rak[i] - 1];
    while (i + k <= n && j + k <= n && s[i + k] == s[j + k]) k++;
    h[rak[i]] = k;
  }
}
void init(long long size) {
  for (long long i = 2; i <= size; i++) aux[i] = aux[i >> 1] + 1;
}
void builder(long long n) {
  for (long long i = 1; i <= n; i++) st[0][i] = h[i];
  for (long long i = 1; i <= 30; i++)
    for (long long r = 1; r + (1 << i) - 1 <= n; r++)
      st[i][r] = min(st[i - 1][r], st[i - 1][r + (1 << i - 1)]);
}
long long lcp(long long a, long long b) {
  if (a == b) return n - a + 1;
  long long x = rak[a], y = rak[b];
  if (x > y) swap(x, y);
  ++x;
  long long t = aux[y - x + 1];
  return min(st[t][x], st[t][y - (1 << t) + 1]);
}
bool cmp(zero a, zero b) {
  return (rak[a.x] == rak[b.x]) ? (a.y < b.y) : (rak[a.x] < rak[b.x]);
}
long long sta[500005], size = 0, ans = 0;
long long num[500005], len1, len2, k;
signed main() {
  init(500005 - 10);
  scanf("%lld%lld", &n, &k);
  scanf("%s", s + 1);
  n = strlen(s + 1);
  work();
  builder(n);
  long long tot = 0, tp = 0, len = 0;
  while (k--) {
    long long n1, n2, nn;
    scanf("%lld%lld", &n1, &n2);
    for (long long i = 1; i <= n1; i++) {
      scanf("%lld", &lib[i].x);
      lib[i].y = 0;
    }
    for (long long i = 1; i <= n2; i++) {
      scanf("%lld", &lib[i + n1].x);
      lib[i + n1].y = 1;
    }
    nn = n1 + n2;
    sort(lib + 1, lib + nn + 1, cmp);
    ans = size = tot = 0;
    for (long long i = 1; i <= nn; i++) {
      tp = 0, len = lcp(lib[i - 1].x, lib[i].x);
      if (lib[i - 1].y == 0) tp++, tot += len;
      while (size > 0 && sta[size] >= len) {
        tot -= num[size] * (sta[size] - len);
        tp += num[size];
        size--;
      }
      sta[++size] = len, num[size] = tp;
      if (lib[i].y == 1) ans += tot;
    }
    size = tot = 0;
    for (long long i = 1; i <= nn; i++) {
      tp = 0, len = lcp(lib[i - 1].x, lib[i].x);
      if (lib[i - 1].y == 1) tp++, tot += len;
      while (size > 0 && sta[size] >= len) {
        tot -= num[size] * (sta[size] - len);
        tp += num[size];
        size--;
      }
      sta[++size] = len, num[size] = tp;
      if (lib[i].y == 0) ans += tot;
    }
    printf("%lld\n", ans);
  }
}