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 + 5;
const int LOG = 20;
const long long INF = 1e18;
const long long MOD = 1e9 + 7;
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
long long sq(long long x) { return (x * x) % MOD; }
long long modP(long long a, long long b) {
  return (!b ? 1 : (sq(modP(a, b / 2)) * (b % 2 ? a : 1)) % MOD);
}
int n, p[maxN], pw;
int R[maxN][LOG];
int a[maxN], l[maxN], r[maxN];
stack<int> st;
vector<int> vec[3];
bool cmp(int x, int y) {
  if (R[x][pw - 1] != R[y][pw - 1]) return (R[x][pw - 1] < R[y][pw - 1]);
  int tmp = (1 << (pw - 1));
  if (max(x, y) + tmp > n) return (x > y);
  return (R[x + tmp][pw - 1] < R[y + tmp][pw - 1]);
}
int LCP(int x, int y) {
  if (x == y) return n - x + 1;
  int ans1 = 0;
  for (int i = LOG - 1; i > -1 && x <= n && y <= n; i--) {
    if (R[x][i] == R[y][i]) {
      ans1 += (1 << i);
      x += (1 << i);
      y += (1 << i);
    }
  }
  return ans1;
}
long long solve(int b) {
  sort(vec[b].begin(), vec[b].end(), cmp);
  long long ans = 0;
  int sz = vec[b].size();
  for (int i = 1; i < sz; i++) a[i] = LCP(vec[b][i], vec[b][i - 1]);
  for (int i = 1; i < sz; i++) {
    while (!st.empty() && a[st.top()] > a[i]) st.pop();
    l[i] = (st.empty() ? 0 : st.top());
    st.push(i);
  }
  while (!st.empty()) st.pop();
  for (int i = sz - 1; i > 0; i--) {
    while (!st.empty() && a[st.top()] >= a[i]) st.pop();
    r[i] = (st.empty() ? sz : st.top());
    st.push(i);
  }
  while (!st.empty()) st.pop();
  for (int i = 1; i < sz; i++) {
    ans += 1ll * a[i] * (i - l[i]) * (r[i] - i);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int q;
  cin >> n >> q;
  string s;
  cin >> s;
  for (int i = 1; i <= n; i++) {
    R[i][0] = s[i - 1];
    p[i] = i;
  }
  for (pw = 1; pw < LOG; pw++) {
    sort(p + 1, p + n + 1, cmp);
    R[p[1]][pw] = 1;
    for (int i = 2; i <= n; i++)
      R[p[i]][pw] = R[p[i - 1]][pw] + cmp(p[i - 1], p[i]);
  }
  while (q--) {
    vec[0].clear();
    vec[1].clear();
    vec[2].clear();
    int s1, s2, inp;
    cin >> s1 >> s2;
    while (s1--) {
      cin >> inp;
      vec[0].push_back(inp);
      vec[2].push_back(inp);
    }
    while (s2--) {
      cin >> inp;
      vec[1].push_back(inp);
      vec[2].push_back(inp);
    }
    long long res = solve(0) + solve(1);
    res = solve(2) - res;
    cout << res << "\n";
  }
  return 0;
}