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>
#pragma GCC optimize(3)
#pragma GCC optimize(2)
using namespace std;
struct SA {
  int p[200111];
  int cnt[200111];
  int c[19][200111];
  int logs[200111];
  int rank[200111];
  const int alphabet = 256;
  int n;
  void init() {
    memset(p, 0, sizeof(p));
    memset(cnt, 0, sizeof(cnt));
    memset(c, 0, sizeof(c));
  }
  void sort_cyclic_shifts(string &s) {
    int h, i;
    n = (int)s.size();
    for (i = 2; i < 200111; ++i) logs[i] = logs[i / 2] + 1;
    for (i = 0; i < n; i++) cnt[(int)s[i]]++;
    for (i = 1; i < alphabet; i++) cnt[i] += cnt[i - 1];
    for (i = 0; i < n; i++) p[--cnt[(int)s[i]]] = i;
    c[0][p[0]] = 0;
    int classes = 1;
    for (i = 1; i < n; i++) {
      if (s[p[i]] != s[p[i - 1]]) classes++;
      c[0][p[i]] = classes - 1;
    }
    vector<int> pn(n), cn(n);
    for (h = 0; (1 << h) < n; ++h) {
      for (i = 0; i < n; i++) {
        pn[i] = p[i] - (1 << h);
        if (pn[i] < 0) pn[i] += n;
      }
      fill(cnt, cnt + classes, 0);
      for (i = 0; i < n; i++) cnt[c[h][pn[i]]]++;
      for (i = 1; i < classes; i++) cnt[i] += cnt[i - 1];
      for (i = n - 1; i >= 0; i--) p[--cnt[c[h][pn[i]]]] = pn[i];
      cn[p[0]] = 0;
      classes = 1;
      for (i = 1; i < n; i++) {
        pair<int, int> cur = {c[h][p[i]], c[h][(p[i] + (1 << h)) % n]};
        pair<int, int> prev = {c[h][p[i - 1]], c[h][(p[i - 1] + (1 << h)) % n]};
        if (cur != prev) ++classes;
        cn[p[i]] = classes - 1;
      }
      for (i = 0; i < n; ++i) {
        c[h + 1][i] = cn[i];
      }
    }
    for (i = 0; i < n; ++i) {
      rank[p[i]] = i;
    }
  }
  int lcp(int i, int j) {
    int ans = 0, k;
    for (k = logs[n]; k >= 0; k--) {
      if (c[k][i] == c[k][j]) {
        ans += 1 << k;
        i += 1 << k;
        j += 1 << k;
      }
    }
    return ans;
  }
};
SA sa;
char S[200111];
string s;
int N, Q;
vector<int> a, b;
vector<pair<int, int>> evt;
static bool cmp(const int &i1, const int &i2) {
  return sa.rank[i1] < sa.rank[i2];
}
void solve() {
  evt.clear();
  sort(a.begin(), a.end(), cmp);
  sort(b.begin(), b.end(), cmp);
  long long ans = 0, curr = 0;
  int i, j, sz1 = a.size(), sz2 = b.size();
  for (i = 0, j = 0; i < sz1; ++i) {
    while (j < sz2 && sa.rank[b[j]] <= sa.rank[a[i]]) {
      evt.emplace_back(b[j++], 0);
    }
    evt.emplace_back(a[i], 1);
  }
  while (j < sz2) evt.emplace_back(b[j++], 0);
  stack<pair<int, int>> stk;
  for (i = 0; i < sz1 + sz2; ++i) {
    if (!evt[i].second) ans += curr;
    if (i != sz1 + sz2 - 1) {
      int p = sa.lcp(evt[i].first, evt[i + 1].first);
      int tol = evt[i].second;
      while (!stk.empty() && stk.top().first >= p) {
        curr -= 1ll * stk.top().first * stk.top().second;
        tol += stk.top().second;
        stk.pop();
      }
      curr += 1ll * p * tol;
      stk.push(make_pair(p, tol));
    }
  }
  while (!stk.empty()) stk.pop();
  evt.clear();
  for (i = 0, j = 0; i < sz1; ++i) {
    while (j < sz2 && sa.rank[b[j]] < sa.rank[a[i]]) {
      evt.emplace_back(b[j++], 0);
    }
    evt.emplace_back(a[i], 1);
  }
  while (j < sz2) evt.emplace_back(b[j++], 0);
  curr = 0;
  for (i = sz1 + sz2 - 1; i >= 0; --i) {
    if (!evt[i].second) ans += curr;
    if (i != 0) {
      int p = sa.lcp(evt[i].first, evt[i - 1].first);
      int tol = evt[i].second;
      while (!stk.empty() && stk.top().first >= p) {
        curr -= 1ll * stk.top().first * stk.top().second;
        tol += stk.top().second;
        stk.pop();
      }
      curr += 1ll * p * tol;
      stk.push(make_pair(p, tol));
    }
  }
  set<int> stt;
  for (auto v : a) stt.insert(v);
  for (auto v : b) {
    if (stt.find(v) != stt.end()) ans += N - v;
  }
  cout << ans << endl;
}
int main() {
  scanf("%d%d", &N, &Q);
  scanf("%s", S);
  s = string(S);
  s += '$';
  sa.sort_cyclic_shifts(s);
  while (Q-- > 0) {
    a.clear(), b.clear();
    int k1, k2, v;
    scanf("%d%d", &k1, &k2);
    while (k1-- > 0) {
      scanf("%d", &v);
      a.push_back(v - 1);
    }
    while (k2-- > 0) {
      scanf("%d", &v);
      b.push_back(v - 1);
    }
    solve();
  }
  return 0;
}