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 N = 212345;
struct RMQ {
  int dp[N][20], log2[N];
  void init(int a[], int n) {
    for (int i = 0; i < n; i++) {
      dp[i][0] = a[i];
    }
    for (int j = 1; (1 << j) <= n; j++) {
      for (int i = 0; i + (1 << j) - 1 < n; i++) {
        dp[i][j] = min(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
      }
    }
    for (int i = 2; i <= n; i++) {
      log2[i] = log2[i - 1];
      if ((i & (i - 1)) == 0) {
        log2[i]++;
      }
    }
  }
  int query(int l, int r) {
    if (l > r) {
      return N;
    }
    int k = log2[r - l + 1];
    return min(dp[l][k], dp[r - (1 << k) + 1][k]);
  }
} rmq;
struct SuffixArray {
  int n;
  char s[N];
  int sa[N], t[N], t1[N], c[N], height[N], rank[N];
  void build(int m) {
    int *x = t, *y = t1;
    for (int i = 0; i < m; i++) {
      c[i] = 0;
    }
    for (int i = 0; i < n; i++) {
      c[x[i] = s[i]]++;
    }
    for (int i = 1; i < m; i++) {
      c[i] += c[i - 1];
    }
    for (int i = n - 1; i >= 0; i--) {
      sa[--c[x[i]]] = i;
    }
    for (int k = 1; k <= n; k <<= 1) {
      int p = 0;
      for (int i = n - k; i < n; i++) {
        y[p++] = i;
      }
      for (int i = 0; i < n; i++) {
        if (sa[i] >= k) {
          y[p++] = sa[i] - k;
        }
      }
      for (int i = 0; i < m; i++) {
        c[i] = 0;
      }
      for (int i = 0; i < n; i++) {
        c[x[y[i]]]++;
      }
      for (int i = 1; i < m; i++) {
        c[i] += c[i - 1];
      }
      for (int i = n - 1; i >= 0; i--) {
        sa[--c[x[y[i]]]] = y[i];
      }
      swap(x, y);
      p = 1, x[sa[0]] = 0;
      for (int i = 1; i < n; i++) {
        x[sa[i]] = y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k]
                       ? p - 1
                       : p++;
      }
      if (p >= n) {
        break;
      }
      m = p;
    }
  }
  void print_sa() {
    for (int i = 0; i < n; i++) {
      cout << sa[i] << " ";
    }
    cout << endl;
    for (int i = 0; i < n; i++) {
      cout << height[i] << " ";
    }
    cout << endl;
  }
  void get() {
    for (int i = 0; i < n; i++) {
      rank[sa[i]] = i;
    }
    int k = 0;
    for (int i = 0; i < n; i++) {
      if (k) {
        k--;
      }
      if (rank[i] == 0) {
        continue;
      }
      int j = sa[rank[i] - 1];
      while (i + k < n && j + k < n && s[i + k] == s[j + k]) {
        k++;
      }
      height[rank[i]] = k;
    }
  }
} sa;
int seq_tail;
pair<int, int> seq[N];
long long solve(vector<int> &a, vector<int> &b) {
  vector<pair<int, int>> order;
  int i = 0, j = 0;
  while (i < a.size() || j < b.size()) {
    if (j == b.size() || i < a.size() && a[i] < b[j]) {
      order.push_back(make_pair(a[i], 0));
      i++;
    } else {
      order.push_back(make_pair(b[j], 1));
      j++;
    }
  }
  seq_tail = 0;
  long long ans = 0, sum = 0;
  for (int i = 0; i < order.size(); i++) {
    if (order[i].second == 0) {
      if (i == order.size() - 1) {
        break;
      }
      seq[seq_tail++] =
          make_pair(rmq.query(order[i].first + 1, order[i + 1].first), 1);
      sum += seq[seq_tail - 1].first;
      while (seq_tail >= 2 &&
             seq[seq_tail - 2].first >= seq[seq_tail - 1].first) {
        sum -= 1ll * (seq[seq_tail - 2].first - seq[seq_tail - 1].first) *
               seq[seq_tail - 2].second;
        seq[seq_tail - 1].second += seq[seq_tail - 2].second;
        seq[seq_tail - 2] = seq[seq_tail - 1];
        seq_tail--;
      }
    } else {
      ans += sum;
      if (i == order.size() - 1) {
        break;
      }
      sum -= 1ll *
             max(0, seq[seq_tail - 1].first -
                        rmq.query(order[i].first + 1, order[i + 1].first)) *
             seq[seq_tail - 1].second;
      seq[seq_tail - 1].first =
          min(seq[seq_tail - 1].first,
              rmq.query(order[i].first + 1, order[i + 1].first));
      while (seq_tail >= 2 &&
             seq[seq_tail - 2].first >= seq[seq_tail - 1].first) {
        sum -= 1ll * (seq[seq_tail - 2].first - seq[seq_tail - 1].first) *
               seq[seq_tail - 2].second;
        seq[seq_tail - 1].second += seq[seq_tail - 2].second;
        seq[seq_tail - 2] = seq[seq_tail - 1];
        seq_tail--;
      }
    }
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  int n, q;
  cin >> n >> q >> sa.s;
  sa.s[n] = 127;
  sa.s[n + 1] = 0;
  sa.n = strlen(sa.s);
  sa.build(128);
  sa.get();
  rmq.init(sa.height, n);
  for (int i = 0; i < q; i++) {
    int la, lb, x;
    cin >> la >> lb;
    vector<int> a, b;
    for (int i = 0; i < la; i++) {
      cin >> x;
      a.push_back(sa.rank[x - 1]);
    }
    for (int i = 0; i < lb; i++) {
      cin >> x;
      b.push_back(sa.rank[x - 1]);
    }
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());
    long long ans = solve(a, b) + solve(b, a);
    for (auto &x : a) {
      if (lower_bound(b.begin(), b.end(), x) !=
          upper_bound(b.begin(), b.end(), x)) {
        ans += n - sa.sa[x];
      }
    }
    cout << ans << endl;
  }
}