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 MOD = 1e9 + 7;
const double PI = acos(-1.);
const double eps = 1e-10;
const int INF = 0x3f3f3f3f;
const long long INFLL = 0x3f3f3f3f3f3f3f3f;
const int MAXN = 1e6 + 5;
char s[MAXN];
int sa[MAXN], Rank[MAXN], height[MAXN];
int wa[MAXN], wb[MAXN], wv[MAXN], wd[MAXN];
int cmp(int *r, int a, int b, int l) {
  return r[a] == r[b] && r[a + l] == r[b + l];
}
void da(char *r, int n, int m = 128) {
  int i, j, p, *x = wa, *y = wb, *t;
  for (i = 0; i < m; i++) wd[i] = 0;
  for (i = 0; i < n; i++) wd[x[i] = r[i]]++;
  for (i = 1; i < m; i++) wd[i] += wd[i - 1];
  for (i = n - 1; i >= 0; i--) sa[--wd[x[i]]] = i;
  for (j = 1, p = 1; p < n; j *= 2, m = p) {
    for (p = 0, i = n - j; i < n; i++) y[p++] = i;
    for (i = 0; i < n; i++)
      if (sa[i] >= j) y[p++] = sa[i] - j;
    for (i = 0; i < n; i++) wv[i] = x[y[i]];
    for (i = 0; i < m; i++) wd[i] = 0;
    for (i = 0; i < n; i++) wd[wv[i]]++;
    for (i = 1; i < m; i++) wd[i] += wd[i - 1];
    for (i = n - 1; i >= 0; i--) sa[--wd[wv[i]]] = y[i];
    for (t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; i++) {
      x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;
    }
  }
}
void calHeight(char *r, int n) {
  int i, j, k = 0;
  for (i = 1; i <= n; i++) Rank[sa[i]] = i;
  for (i = 0; i < n; height[Rank[i++]] = k) {
    for (k ? k-- : 0, j = sa[Rank[i] - 1]; r[i + k] == r[j + k]; k++)
      ;
  }
}
int d[MAXN][20];
void RMQ_init(int a[], int n) {
  for (int i = 0; i < n; i++) d[i][0] = a[i];
  for (int j = 1; (1 << j) <= n; j++)
    for (int i = 0; i + (1 << j) - 1 < n; i++)
      d[i][j] = min(d[i][j - 1], d[i + (1 << (j - 1))][j - 1]);
}
int n, q;
int RMQ(int L, int R) {
  int k = 0;
  if (L > R) swap(L, R);
  if (L == R) return n - sa[L];
  L++;
  while ((1 << (k + 1)) <= R - L + 1) k++;
  return min(d[L][k], d[R - (1 << k) + 1][k]);
}
pair<int, int> idx[MAXN];
long long solve(int len) {
  stack<pair<int, int> > sk;
  long long sum = 0, ans = 0;
  for (int i = 1; i <= len; i++) {
    int id = idx[i].first;
    if (i > 1) {
      int h = RMQ(id, idx[i - 1].first);
      int num = 0;
      while (sk.size() && sk.top().first >= h) {
        pair<int, int> now = sk.top();
        sk.pop();
        sum -= (long long)now.first * now.second;
        num += now.second;
      }
      sk.push(make_pair(h, num));
      sum += (long long)h * num;
    }
    if (idx[i].second == 1)
      ans += sum;
    else
      sk.push(make_pair(n - sa[id], 1)), sum += n - sa[id];
  }
  return ans;
}
int main() {
  scanf("%d %d %s", &n, &q, s);
  da(s, n + 1);
  calHeight(s, n);
  RMQ_init(height, n + 1);
  while (q--) {
    int k, l;
    scanf("%d %d", &k, &l);
    for (int i = 1; i <= k; i++) {
      scanf("%d", &idx[i].first);
      idx[i].first = Rank[idx[i].first - 1];
      idx[i].second = 1;
    }
    for (int i = k + 1; i <= k + l; i++) {
      scanf("%d", &idx[i].first);
      idx[i].first = Rank[idx[i].first - 1];
      idx[i].second = -1;
    }
    int len = l + k;
    sort(idx + 1, idx + 1 + len);
    long long ans = 0;
    ans += solve(len);
    reverse(idx + 1, idx + 1 + len);
    ans += solve(len);
    printf("%lld\n", ans);
  }
  return 0;
}