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 maxn = 4e5 + 7;
char s[maxn];
int sa[maxn], str[maxn], _rank[maxn], height[maxn];
int t1[maxn], t2[maxn], c[maxn];
int RMQ[maxn], mm[maxn], best[20][maxn], len;
bool cmp(int *r, int a, int b, int l) {
  return r[a] == r[b] && r[a + l] == r[b + l];
}
void da(int n, int m) {
  int i, j, p, *x = t1, *y = t2;
  for (i = 0; i < m; i++) c[i] = 0;
  for (i = 0; i < n; i++) c[x[i] = str[i]]++;
  for (i = 1; i < m; i++) c[i] += c[i - 1];
  for (i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;
  for (j = 1; j <= n; j <<= 1) {
    p = 0;
    for (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 < m; i++) c[i] = 0;
    for (i = 0; i < n; i++) c[x[y[i]]]++;
    for (i = 1; i < m; i++) c[i] += c[i - 1];
    for (i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
    swap(x, y);
    p = 1;
    x[sa[0]] = 0;
    for (i = 1; i < n; i++)
      x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;
    if (p >= n) break;
    m = p;
  }
  int k = 0;
  n--;
  for (i = 0; i <= n; i++) _rank[sa[i]] = i;
  for (i = 0; i < n; i++) {
    if (k) k--;
    j = sa[_rank[i] - 1];
    while (str[i + k] == str[j + k]) k++;
    height[_rank[i]] = k;
  }
}
void initRMQ(int n) {
  mm[0] = -1;
  for (int i = 1; i <= n; i++)
    mm[i] = ((i & (i - 1)) == 0) ? mm[i - 1] + 1 : mm[i - 1];
  for (int i = 1; i <= n; i++) best[0][i] = i;
  for (int i = 1; i <= mm[n]; i++)
    for (int j = 1; j + (1 << i) - 1 <= n; j++) {
      int a = best[i - 1][j];
      int b = best[i - 1][j + (1 << (i - 1))];
      if (RMQ[a] < RMQ[b])
        best[i][j] = a;
      else
        best[i][j] = b;
    }
}
int askRMQ(int a, int b) {
  int t = mm[b - a + 1];
  b -= (1 << t) - 1;
  a = best[t][a];
  b = best[t][b];
  return RMQ[a] < RMQ[b] ? a : b;
}
int lcp(int a, int b) {
  if (a == b) return len - a;
  a = _rank[a];
  b = _rank[b];
  if (a > b) swap(a, b);
  return height[askRMQ(a + 1, b)];
}
int _stack[maxn][2];
int main() {
  int q, lim, x;
  scanf("%d%d", &len, &q);
  scanf("%s", s);
  for (int i = 0; i < len; i++) str[i] = s[i];
  str[len] = 0;
  da(len + 1, 256);
  for (int i = 1; i <= len; i++) RMQ[i] = height[i];
  initRMQ(len);
  while (q--) {
    int n, m;
    long long ans = 0;
    vector<pair<int, int> > Q;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d", &x), Q.emplace_back(x - 1, 1);
    for (int i = 1; i <= m; i++) scanf("%d", &x), Q.emplace_back(x - 1, 0);
    sort(Q.begin(), Q.end(),
         [](const pair<int, int> &L, const pair<int, int> &R) {
           if (L.first == R.first) return L.second < R.second;
           int l = lcp(L.first, R.first);
           return s[L.first + l] < s[R.first + l];
         });
    int sz = Q.size();
    for (int ss = 0; ss < 2; ss++) {
      long long tot = 0;
      map<int, int> W;
      for (int i = 0; i < sz; i++) {
        int id = Q[i].first;
        int state = Q[i].second;
        if (state == 0)
          ans += tot;
        else
          W[len - id]++, tot += len - id;
        if (i < sz - 1) {
          int LCP = lcp(Q[i].first, Q[i + 1].first);
          while (!W.empty()) {
            auto it = --W.end();
            if (it->first <= LCP) break;
            tot -= 1LL * (it->first - LCP) * it->second;
            W[LCP] += it->second;
            W.erase(it);
          }
        }
      }
      reverse(Q.begin(), Q.end());
    }
    printf("%lld\n", ans);
  }
  return 0;
}