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 + 20;
string s;
int n, gap, sa[maxn], pos[maxn], tmp[maxn], lcp[maxn], Mn[maxn * 4];
inline bool sufCmp(int i, int j) {
  if (pos[i] != pos[j]) return pos[i] < pos[j];
  i += gap;
  j += gap;
  return (i < n && j < n) ? pos[i] < pos[j] : i > j;
}
inline void buildSA() {
  for (int i = 0; i < n; i++) sa[i] = i, pos[i] = s[i];
  for (gap = 1;; gap *= 2) {
    sort(sa, sa + n, sufCmp);
    for (int i = 0; i < n - 1; i++)
      tmp[i + 1] = tmp[i] + sufCmp(sa[i], sa[i + 1]);
    for (int i = 0; i < n; i++) pos[sa[i]] = tmp[i];
    if (tmp[n - 1] == n - 1) break;
  }
}
inline void buildLCP() {
  for (int i = 0, k = 0; i < n; i++)
    if (pos[i] != n - 1) {
      for (int j = sa[pos[i] + 1]; k < n && s[i + k] == s[j + k];) k++;
      lcp[pos[i]] = k;
      if (k) k--;
    }
}
void buildSeg(int s = 0, int e = n - 1, int v = 1) {
  if (e - s < 2) {
    Mn[v] = lcp[s];
    return;
  }
  int m = (s + e) / 2;
  buildSeg(s, m, 2 * v);
  buildSeg(m, e, 2 * v + 1);
  Mn[v] = min(Mn[2 * v], Mn[2 * v + 1]);
}
int GetLcp(int l, int r, int s = 0, int e = n - 1, int v = 1) {
  if (l == r) return n - sa[l];
  if (l <= s && e <= r) return Mn[v];
  if (r <= s || e <= l) return maxn;
  int m = (s + e) / 2;
  return min(GetLcp(l, r, s, m, 2 * v), GetLcp(l, r, m, e, 2 * v + 1));
}
int L[maxn], R[maxn];
long long get(vector<int> x) {
  for (auto &y : x) y = pos[y];
  sort(x.begin(), x.end());
  int m = x.size();
  m--;
  for (int i = 0; i < m; i++) tmp[i] = GetLcp(x[i], x[i + 1]);
  stack<int> st;
  for (int i = 0; i < m; i++) {
    while (!st.empty() && tmp[st.top()] >= tmp[i]) st.pop();
    if (st.empty())
      L[i] = -1;
    else
      L[i] = st.top();
    st.push(i);
  }
  while (!st.empty()) st.pop();
  for (int i = m - 1; i >= 0; i--) {
    while (!st.empty() && tmp[st.top()] > tmp[i]) st.pop();
    if (st.empty())
      R[i] = m;
    else
      R[i] = st.top();
    st.push(i);
  }
  long long res = 0;
  for (int i = 0; i < m; i++) res += 1LL * (R[i] - i) * (i - L[i]) * tmp[i];
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int q;
  cin >> n >> q >> s;
  buildSA();
  buildLCP();
  buildSeg();
  while (q--) {
    int k, l;
    cin >> k >> l;
    vector<int> A, B, C;
    while (k--) {
      int x;
      cin >> x;
      A.push_back(x - 1);
      C.push_back(x - 1);
    }
    while (l--) {
      int x;
      cin >> x;
      B.push_back(x - 1);
      C.push_back(x - 1);
    }
    cout << get(C) - get(B) - get(A) << endl;
  }
}