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 = 200010;
const int LOG = 20;
int n, q;
char s[N];
int sa[N], tmp[N], pos[N];
int len, cnt[N], lcp[N];
int rmq[LOG][N];
bool cmp(int u, int v) {
  if (pos[u] != pos[v]) {
    return pos[u] < pos[v];
  }
  return (u + len <= n && v + len <= n) ? pos[u + len] < pos[v + len] : u > v;
}
void radix(int delta) {
  memset(cnt, 0, sizeof cnt);
  for (int i = 1; i <= n; i++) {
    cnt[i + delta <= n ? pos[i + delta] : 0]++;
  }
  for (int i = 1; i < N; i++) {
    cnt[i] += cnt[i - 1];
  }
  for (int i = n; i > 0; i--) {
    int id = sa[i];
    tmp[cnt[id + delta <= n ? pos[id + delta] : 0]--] = id;
  }
  for (int i = 1; i <= n; i++) {
    sa[i] = tmp[i];
  }
}
void buildSA() {
  for (int i = 1; i <= n; i++) {
    sa[i] = i;
    pos[i] = s[i];
  }
  len = 1;
  while (1) {
    radix(len);
    radix(0);
    tmp[1] = 1;
    for (int i = 2; i <= n; i++) {
      tmp[i] = tmp[i - 1] + cmp(sa[i - 1], sa[i]);
    }
    for (int i = 1; i <= n; i++) {
      pos[sa[i]] = tmp[i];
    }
    if (tmp[n] == n) {
      break;
    }
    len <<= 1;
  }
  len = 0;
  for (int i = 1; i <= n; i++) {
    if (pos[i] == n) {
      continue;
    }
    int j = sa[pos[i] + 1];
    while (s[i + len] == s[j + len]) {
      len++;
    }
    lcp[pos[i]] = len;
    if (len) {
      len--;
    }
  }
  for (int i = 1; i < n; i++) {
    rmq[0][i] = lcp[i];
  }
  for (int i = 1; i < LOG; i++) {
    for (int j = 1; j + (1 << i) - 1 < n; j++) {
      rmq[i][j] = min(rmq[i - 1][j], rmq[i - 1][j + (1 << (i - 1))]);
    }
  }
}
int getVal(int l, int r) {
  int lg = log2(r - l + 1);
  return min(rmq[lg][l], rmq[lg][r - (1 << lg) + 1]);
}
void compress(vector<pair<int, int> > &st, long long &curRes, int val) {
  pair<int, int> foo = {0, val};
  while (!st.empty()) {
    pair<int, int> now = st.back();
    if (now.second >= foo.second) {
      curRes -= 1ll * now.first * now.second;
      foo.first += now.first;
      st.pop_back();
    } else {
      break;
    }
  }
  if (foo.first > 0) {
    st.push_back(foo);
    curRes += 1ll * foo.first * foo.second;
  }
}
long long go(vector<int> &a, vector<int> &b, bool f) {
  int cur = 0;
  long long res = 0;
  long long curRes = 0;
  vector<pair<int, int> > st;
  for (int i = 0; i < b.size(); i++) {
    if (!st.empty()) {
      assert(cur > 0);
      int foo = (a[cur - 1] == b[i] ? n - sa[a[cur - 1]] + 1
                                    : getVal(a[cur - 1], b[i] - 1));
      compress(st, curRes, foo);
    }
    while (cur < a.size() && a[cur] < b[i]) {
      int foo = getVal(a[cur], b[i] - 1);
      st.push_back({1, foo});
      curRes += foo;
      compress(st, curRes, foo);
      cur++;
    }
    if (f && cur < a.size() && a[cur] == b[i]) {
      int foo = n - sa[a[cur]] + 1;
      st.push_back({1, foo});
      curRes += foo;
      compress(st, curRes, foo);
      cur++;
    }
    res += curRes;
  }
  return res;
}
int main() {
  scanf("%d %d", &n, &q);
  scanf("%s", s + 1);
  buildSA();
  while (q--) {
    int m[2];
    vector<int> a[2];
    cin >> m[0] >> m[1];
    for (int id = 0; id < 2; id++) {
      for (int i = 1; i <= m[id]; i++) {
        int u;
        cin >> u;
        a[id].push_back(pos[u]);
      }
    }
    sort(a[0].begin(), a[0].end());
    sort(a[1].begin(), a[1].end());
    long long res = 0;
    res += go(a[0], a[1], 0);
    res += go(a[1], a[0], 1);
    printf("%lld\n", res);
  }
  return 0;
}