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 long long N = 500005;
struct ab {
  long long x, y;
} a[N];
char s[N];
long long rak[N], tp[N], sa[N], t[N], h[N], dp[N][2], lg[N], st[N][31], n,
    m = 180;
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 = 180;
  for (long long i = 1; i <= n; i++) rak[i] = s[i] - '0' + 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 rmq() {
  for (long long i = 1; i <= n; i++) {
    st[i][0] = h[i];
  }
  for (long long i = 1; i <= 30; i++) {
    for (long long j = 1; j + (1 << i) - 1 <= n; j++) {
      st[j][i] = min(st[j][i - 1], st[j + (1 << (i - 1))][i - 1]);
    }
  }
}
long long lcp(long long x, long long y) {
  if (x == y) return n - x + 1;
  x = rak[x], y = rak[y];
  if (x > y) swap(x, y);
  x++;
  long long z = lg[y - x + 1];
  return min(st[x][z], st[y - (1 << z) + 1][z]);
}
long long cmp(ab x, ab y) {
  return (rak[x.x] == rak[y.x]) ? (x.y < y.y) : (rak[x.x] < rak[y.x]);
}
signed main() {
  for (long long i = 2; i < N; i++) lg[i] = lg[i >> 1] + 1;
  long long q;
  scanf("%lld%lld%s", &n, &q, s + 1);
  work();
  rmq();
  while (q--) {
    long long n1, n2;
    scanf("%lld%lld", &n1, &n2);
    for (long long i = 1; i <= n1; i++) scanf("%lld", &a[i].x), a[i].y = 0;
    for (long long i = 1; i <= n2; i++)
      scanf("%lld", &a[i + n1].x), a[i + n1].y = 1;
    long long nn = n1 + n2;
    sort(a + 1, a + 1 + nn, cmp);
    long long top = 0, sum = 0, ans = 0;
    for (long long i = 1; i <= nn; i++) {
      long long tot = 0, len = lcp(a[i - 1].x, a[i].x);
      if (a[i - 1].y == 0) tot++, sum += len;
      while (top && dp[top][0] >= len) {
        tot += dp[top][1];
        sum -= dp[top][1] * (dp[top][0] - len);
        top--;
      }
      dp[++top][0] = len, dp[top][1] = tot;
      if (a[i].y == 1) ans += sum;
    }
    top = sum = 0;
    for (long long i = 1; i <= nn; i++) {
      long long tot = 0, len = lcp(a[i - 1].x, a[i].x);
      if (a[i - 1].y == 1) tot++, sum += len;
      while (top && dp[top][0] >= len) {
        tot += dp[top][1];
        sum -= dp[top][1] * (dp[top][0] - len);
        top--;
      }
      dp[++top][0] = len, dp[top][1] = tot;
      if (a[i].y == 0) ans += sum;
    }
    printf("%lld\n", ans);
  }
  return 0;
}