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 = 2e5 + 10;
long long n, m;
long long T;
long long lenL, lenR;
long long a[N], b[N], xv[N << 1], hh[N << 1], lg[N], st[N << 1][23];
char s[N];
long long rnk[N], tp[N], height[N], sa[N], to[N], top;
long long sta[N << 1], l[N << 1], r[N << 1];
long long chang[N];
void RS() {
  for (long long i = 1; i <= m; i++) to[i] = 0;
  for (long long i = 1; i <= n; i++) to[rnk[i]]++;
  for (long long i = 1; i <= m; i++) to[i] += to[i - 1];
  for (long long i = n; i >= 1; i--) sa[to[rnk[tp[i]]]--] = tp[i];
}
void SA() {
  for (long long i = 1; i <= n; i++) rnk[i] = s[i], tp[i] = i;
  RS();
  for (long long k = 1; k <= n; k <<= 1) {
    long long p = 0;
    for (long long i = n - k + 1; i <= n; i++) tp[++p] = i;
    for (long long i = 1; i <= n; i++)
      if (sa[i] > k) tp[++p] = sa[i] - k;
    RS();
    swap(rnk, tp);
    rnk[sa[1]] = p = 1;
    for (long long i = 2; i <= n; i++)
      rnk[sa[i]] =
          (tp[sa[i]] == tp[sa[i - 1]] && tp[sa[i] + k] == tp[sa[i - 1] + k])
              ? p
              : ++p;
    if (p == n) break;
    m = p;
  }
  for (long long i = 1, j = 0; i <= n; i++) {
    if (rnk[i] == 1) continue;
    if (j) j--;
    long long tmp = sa[rnk[i] - 1];
    while (tmp + j <= n && i + j <= n && s[tmp + j] == s[i + j]) j++;
    height[rnk[i]] = j;
  }
  for (long long i = 1; i <= n; i++) chang[i] = n - sa[i] + 1;
}
long long cha(long long l, long long r) {
  if (l == r) return chang[l];
  l++;
  long long p = lg[r - l + 1];
  return min(st[l][p], st[r - (1 << p) + 1][p]);
}
long long work(long long p[], long long len) {
  long long res = 0;
  top = 0;
  for (long long i = 2; i <= len; i++) hh[i] = cha(p[i - 1], p[i]);
  for (long long i = 1; i <= len; i++) {
    while (top && hh[sta[top]] >= hh[i]) {
      r[sta[top]] = i - 1;
      top--;
    }
    l[i] = sta[top] + 1;
    sta[++top] = i;
  }
  while (top) r[sta[top--]] = len;
  for (long long i = 1; i <= len; i++)
    res += hh[i] * (i - l[i] + 1) * (r[i] - i + 1);
  return res;
}
signed main() {
  memset(st, 0x3f, sizeof(st));
  cin >> n >> T;
  scanf("%s", s + 1);
  m = 'z';
  SA();
  for (long long i = 1; i <= n; i++) st[i][0] = height[i];
  lg[0] = -1;
  for (long long i = 1; i <= 2e5; i++) lg[i] = lg[i >> 1] + 1;
  for (long long j = 1; j <= 20; j++)
    for (long long i = 1; i + (1 << j) - 1 <= n; i++)
      st[i][j] = min(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
  while (T--) {
    long long ans = 0;
    scanf("%lld%lld", &lenL, &lenR);
    for (long long i = 1; i <= lenL; i++)
      scanf("%lld", &a[i]), xv[i] = rnk[a[i]];
    for (long long i = 1; i <= lenR; i++)
      scanf("%lld", &b[i]), xv[i + lenL] = rnk[b[i]];
    sort(xv + 1, xv + lenL + lenR + 1);
    ans += work(xv, lenL + lenR);
    for (long long i = 1; i <= lenL; i++) xv[i] = rnk[a[i]];
    sort(xv + 1, xv + lenL + 1);
    ans -= work(xv, lenL);
    for (long long i = 1; i <= lenR; i++) xv[i] = rnk[b[i]];
    sort(xv + 1, xv + lenR + 1);
    ans -= work(xv, lenR);
    printf("%lld\n", ans);
  }
}