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 = (int)4e5 + 10;
const int mod = (int)998244353;
const int inf = (1 << 30) - 1;
int n, q;
string s;
int suff[maxn];
int col[maxn];
int head[maxn];
int nsuff[maxn];
int ncol[maxn];
int pos[maxn];
int lcp[maxn];
int lg[maxn];
int t[18][maxn];
bool cmps(int i, int j) { return s[i] < s[j]; }
void suffix_array() {
  s += "#";
  n++;
  for (int i = 0; i < n; i++) {
    suff[i] = i;
  }
  sort(suff, suff + n, cmps);
  int cn = 0;
  for (int i = 0; i < n; i++) {
    if (i == 0 || s[suff[i]] != s[suff[i - 1]]) {
      col[suff[i]] = cn;
      head[cn] = i;
      cn++;
    } else {
      col[suff[i]] = cn - 1;
    }
  }
  for (int len = 1; len <= n; len *= 2) {
    for (int i = 0; i < n; i++) {
      int j = suff[i] - len;
      if (j < 0) j += n;
      int c = col[j];
      nsuff[head[c]++] = j;
    }
    cn = 0;
    for (int i = 0; i < n; i++) {
      if (i == 0 || col[nsuff[i]] != col[nsuff[i - 1]] ||
          col[(nsuff[i] + len) % n] != col[(nsuff[i - 1] + len) % n]) {
        ncol[nsuff[i]] = cn;
        head[cn] = i;
        cn++;
      } else {
        ncol[nsuff[i]] = cn - 1;
      }
    }
    for (int i = 0; i < n; i++) {
      suff[i] = nsuff[i];
      col[i] = ncol[i];
    }
  }
  for (int i = 0; i < n; i++) {
    suff[i] = suff[i + 1];
  }
  n--;
  for (int i = 0; i < n; i++) {
    pos[suff[i]] = i;
  }
  s.pop_back();
  int len = 0;
  for (int i = 0; i < n; i++) {
    int j = pos[i];
    if (j == n - 1) {
      len = 0;
      lcp[j] = 0;
      continue;
    }
    while (suff[j] + len < n && suff[j + 1] + len < n &&
           s[suff[j] + len] == s[suff[j + 1] + len]) {
      len++;
    }
    lcp[j] = len;
    len = max(len - 1, 0);
  }
  for (int i = 0; i < n; i++) {
    t[0][i] = lcp[i];
  }
  lg[0] = -1;
  for (int i = 1; i <= n; i++) {
    lg[i] = lg[i / 2] + 1;
  }
  for (int len = 1; (1 << len) <= n; len++) {
    for (int i = 0; i + (1 << len) - 1 < n; i++) {
      t[len][i] = min(t[len - 1][i], t[len - 1][i + (1 << (len - 1))]);
    }
  }
}
int get_min(int l, int r) {
  if (l == r + 1) {
    return n - suff[l];
  }
  int len = lg[r - l + 1];
  return min(t[len][l], t[len][r - (1 << len) + 1]);
}
int p[maxn];
int get(int v) {
  if (p[v] == v) return v;
  return p[v] = get(p[v]);
}
int sz[maxn][2];
void calc() {
  int x, y;
  vector<pair<int, int> > a;
  cin >> x >> y;
  y += x;
  for (int i = 0; i < y; i++) {
    int z;
    cin >> z;
    z--;
    a.emplace_back(make_pair(pos[z], i < x));
  }
  sort(a.begin(), a.end());
  vector<pair<int, int> > w;
  for (int i = 0; i + 1 < a.size(); i++) {
    w.emplace_back(make_pair(get_min(a[i].first, a[i + 1].first - 1), i));
  }
  sort(w.begin(), w.end());
  for (int i = 0; i < y; i++) {
    p[i] = i;
    sz[i][0] = sz[i][1] = 0;
    sz[i][a[i].second] = 1;
  }
  reverse(w.begin(), w.end());
  long long ans = 0;
  for (int i = 0; i < w.size(); i++) {
    long long d = w[i].first;
    int l = w[i].second;
    int r = l + 1;
    int pa = get(l);
    int pb = get(r);
    ans += d * sz[pa][0] * sz[pb][1];
    ans += d * sz[pa][1] * sz[pb][0];
    p[pa] = pb;
    sz[pb][0] += sz[pa][0];
    sz[pb][1] += sz[pa][1];
  }
  cout << ans << "\n";
}
void solve() {
  cin >> n >> q >> s;
  suffix_array();
  for (int i = 0; i < n; i++) {
  }
  for (int i = 0; i < q; i++) {
    calc();
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}