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[200010];
int sa[200010], rk[200010], height[200010], h[200010];
int t1[200010], t2[200010], c[200010];
int rmq[200010][19], tat[200010];
int n, m;
void build(int m) {
  int *x = t1, *y = t2;
  for (int i = 1; i <= m; i++) c[i] = 0;
  for (int i = 1; i <= n; i++) c[x[i] = s[i]]++;
  for (int i = 1; i <= m; i++) c[i] += c[i - 1];
  for (int i = n; i >= 1; i--) sa[c[x[i]]--] = i;
  for (int k = 1; k <= n; k <<= 1) {
    int p = 0;
    for (int i = n - k + 1; i <= n; i++) y[++p] = i;
    for (int i = 1; i <= n; i++)
      if (sa[i] > k) y[++p] = sa[i] - k;
    for (int i = 1; i <= m; i++) c[i] = 0;
    for (int i = 1; i <= n; i++) c[x[i]]++;
    for (int i = 1; i <= m; i++) c[i] += c[i - 1];
    for (int i = n; i >= 1; i--) sa[c[x[y[i]]]--] = y[i];
    swap(x, y), p = 1, x[sa[1]] = 1;
    for (int i = 2; i <= n; i++)
      x[sa[i]] = (y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k])
                     ? p
                     : ++p;
    if (p >= n) break;
    m = p;
  }
}
int query(int l, int r) {
  return min(rmq[l][tat[r - l + 1]],
             rmq[r - (1 << tat[r - l + 1]) + 1][tat[r - l + 1]]);
}
int lcp(int x, int y) {
  if (x == y) return n - x + 1;
  if (rk[x] > rk[y]) swap(x, y);
  return query(rk[x] + 1, rk[y]);
}
bool comp(const pair<int, int> &a, const pair<int, int> &b) {
  if (a.first != b.first) return rk[a.first] < rk[b.first];
  return a.second < b.second;
}
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s + 1);
  build(2333);
  for (int i = 1; i <= n; i++) rk[sa[i]] = i;
  for (int i = 1; i <= n; i++) {
    h[i] = h[i - 1] - (h[i - 1] > 0);
    while (h[i] <= n && s[i + h[i]] == s[sa[rk[i] - 1] + h[i]]) h[i]++;
    height[rk[i]] = h[i];
  }
  for (int i = 1; i <= n; i++) rmq[i][0] = height[i];
  for (int i = 3; i <= n; i++) tat[i] = tat[(i + 1) / 2] + 1;
  for (int j = 1; j <= 18; j++)
    for (int i = 1; i + (1 << j) - 1 <= n; i++)
      rmq[i][j] = min(rmq[i][j - 1], rmq[i + (1 << (j - 1))][j - 1]);
  for (int x, y, i = 1; i <= m; i++) {
    vector<pair<int, int>> a;
    scanf("%d%d", &x, &y);
    for (int b, j = 1; j <= x; j++) scanf("%d", &b), a.emplace_back(b, 1);
    for (int b, j = 1; j <= y; j++) scanf("%d", &b), a.emplace_back(b, 0);
    sort(a.begin(), a.end(), comp);
    int sz = a.size();
    long long ans = 0;
    for (int sb = 0; sb <= 1; sb++) {
      long long sum = 0;
      map<int, int> cnt;
      for (int i = 0; i < sz; i++) {
        if (a[i].second == 0)
          ans += sum;
        else
          cnt[n - a[i].first + 1]++, sum += n - a[i].first + 1;
        if (i + 1 < sz) {
          int len = lcp(a[i].first, a[i + 1].first);
          while (!cnt.empty()) {
            map<int, int>::iterator t = --cnt.end();
            if (t->first <= len) break;
            sum -= t->first * (long long)t->second;
            cnt[len] += t->second;
            sum += len * (long long)t->second;
            cnt.erase(t);
          }
        }
      }
      reverse(a.begin(), a.end());
    }
    cout << ans << endl;
  }
  return 0;
}