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;
inline int Getint() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) ch != '-' ?: f = -1, ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
const int N = 200005;
int n, Q;
char s[N];
struct node {
  int son[26], len, fail;
} tr[N * 2];
int lst = 1, sz = 1, pos[N];
void insert(int x) {
  int np = ++sz, p = lst;
  lst = np;
  tr[np].len = tr[p].len + 1;
  while (p && !tr[p].son[x]) tr[p].son[x] = np, p = tr[p].fail;
  if (!p)
    tr[np].fail = 1;
  else {
    int q = tr[p].son[x];
    if (tr[q].len == tr[p].len + 1)
      tr[np].fail = q;
    else {
      int nq = ++sz;
      tr[nq] = tr[q];
      tr[nq].len = tr[p].len + 1;
      tr[np].fail = tr[q].fail = nq;
      while (p && tr[p].son[x] == q) tr[p].son[x] = nq, p = tr[p].fail;
    }
  }
}
bool imp[N * 2], imp2[N * 2];
int dfn[N * 2], dep[N * 2], p[N * 2][20];
vector<int> G[N * 2];
void dfs(int x, int fa) {
  dfn[x] = ++dfn[0];
  dep[x] = dep[fa] + 1;
  p[x][0] = fa;
  for (int i = 1; i <= 18; i++) p[x][i] = p[p[x][i - 1]][i - 1];
  for (int y : G[x]) dfs(y, x);
}
int LCA(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (int i = 17; ~i; i--)
    if (dep[p[x][i]] >= dep[y]) x = p[x][i];
  if (x == y) return x;
  for (int i = 17; ~i; i--)
    if (p[x][i] != p[y][i]) x = p[x][i], y = p[y][i];
  return p[x][0];
}
bool cmp(int x, int y) { return dfn[x] < dfn[y]; }
int cnt, h[N * 2];
struct edge {
  int to, next;
} e[N * 2];
void addedge(int x, int y) {
  e[++cnt] = (edge){y, h[x]};
  h[x] = cnt;
}
int size[N * 2], size2[N * 2];
long long ans;
void dp(int x) {
  size[x] = imp[x];
  size2[x] = imp2[x];
  if (imp[x] && imp2[x]) ans += tr[x].len;
  for (int& i = h[x]; i; i = e[i].next) {
    int y = e[i].to;
    dp(y);
    ans += 1ll * size[y] * size2[x] * tr[x].len;
    ans += 1ll * size2[y] * size[x] * tr[x].len;
    size[x] += size[y];
    size2[x] += size2[y];
  }
  imp[x] = imp2[x] = 0;
}
int main() {
  n = Getint();
  Q = Getint();
  scanf("%s", s + 1);
  for (int i = n; i; i--) insert(s[i] - 'a');
  for (int i = n, now = 1; i; i--) {
    now = tr[now].son[s[i] - 'a'];
    pos[i] = now;
  }
  for (int i = 2; i <= sz; i++) G[tr[i].fail].push_back(i);
  dfs(1, 0);
  while (Q--) {
    int k = Getint(), l = Getint();
    vector<int> a;
    for (int i = 1; i <= k; i++) {
      int x = pos[Getint()];
      a.push_back(x);
      imp[x] = 1;
    }
    for (int i = 1; i <= l; i++) {
      int x = pos[Getint()];
      a.push_back(x);
      imp2[x] = 1;
    }
    sort(a.begin(), a.end());
    a.erase(unique(a.begin(), a.end()), a.end());
    sort(a.begin(), a.end(), cmp);
    vector<int> st;
    cnt = 0;
    for (int now : a) {
      int f = 0;
      while (st.size()) {
        f = LCA(now, st.back());
        if (st.size() > 1 && dep[f] < dep[st[st.size() - 2]]) {
          addedge(st[st.size() - 2], st.back());
          st.pop_back();
        } else {
          if (dep[f] < dep[st.back()]) {
            addedge(f, st.back());
            st.pop_back();
          }
          break;
        }
      }
      if (f && (!st.size() || st.back() != f)) st.push_back(f);
      st.push_back(now);
    }
    while (st.size() > 1) addedge(st[st.size() - 2], st.back()), st.pop_back();
    ans = 0;
    dp(st[0]);
    cout << ans << "\n";
  }
}