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("inline", 3)
using namespace std;
const int N = 8e5 + 5, B = 20;
char s[N];
int n, fa[N], len[N], sz[N], a[N], cc[N], np = 1, cnt = 1;
map<int, int> c[N];
int poi[N];
vector<int> e[N];
int st[B][N << 1], dfn[N], pos[N], tim;
int qq[N << 1], pre[N], za[N], zb[N], siz;
bool cmp(int a, int b) { return dfn[a] < dfn[b]; }
void dfs(int u) {
  dfn[u] = ++tim;
  pos[tim] = u;
  st[0][tim] = dfn[u];
  for (auto v : e[u]) dfs(v), st[0][++tim] = dfn[u];
}
void bui_st() {
  for (int i = 1; i <= B - 1; ++i)
    for (int j = 1; j + (1 << i) - 1 <= tim; ++j)
      st[i][j] = min(st[i - 1][j], st[i - 1][j + (1 << i - 1)]);
}
int lca(int a, int b) {
  if ((a = dfn[a]) > (b = dfn[b])) swap(a, b);
  int d = floor(log2(b - a + 1));
  return pos[min(st[d][a], st[d][b - (1 << d) + 1])];
}
void append(int x) {
  int p = np;
  np = ++cnt;
  len[np] = len[p] + 1;
  sz[np] = 1;
  for (; p && !c[p][x]; p = fa[p]) c[p][x] = np;
  if (!p)
    fa[np] = 1;
  else {
    int q = c[p][x];
    if (len[q] == len[p] + 1)
      fa[np] = q;
    else {
      int nq = ++cnt;
      len[nq] = len[p] + 1;
      c[nq] = c[q];
      fa[nq] = fa[q];
      fa[q] = fa[np] = nq;
      for (; c[p][x] == q; p = fa[p]) c[p][x] = nq;
    }
  }
}
void get_edge() {
  for (int i = 1; i <= cnt; ++i) e[fa[i]].push_back(i);
}
void ini2() {
  while (siz) za[qq[siz]] = zb[qq[siz]] = 0, --siz;
  int la, lb, x;
  scanf("%d%d", &la, &lb);
  for (int i = 1; i <= la; ++i)
    scanf("%d", &x), ++za[qq[++siz] = poi[n + 1 - x]];
  for (int i = 1; i <= lb; ++i)
    scanf("%d", &x), ++zb[qq[++siz] = poi[n + 1 - x]];
  sort(qq + 1, qq + siz + 1, cmp);
  for (int i = siz; i >= 2; --i) qq[++siz] = lca(qq[i], qq[i - 1]);
  sort(qq + 1, qq + siz + 1, cmp);
  siz = unique(qq + 1, qq + siz + 1) - qq - 1;
  stack<int> Q;
  Q.push(qq[1]);
  pre[qq[1]] = 0;
  for (int i = 2; i <= siz; ++i) {
    for (int t; t = Q.top(), lca(t, qq[i]) != t;) Q.pop();
    pre[qq[i]] = Q.top();
    Q.push(qq[i]);
  }
}
long long solve() {
  long long re = 0;
  for (int i = siz; i >= 1; --i) {
    int x = qq[i];
    re += 1ll * za[x] * zb[x] * (len[x] - len[pre[x]]);
    za[pre[x]] += za[x];
    zb[pre[x]] += zb[x];
  }
  return re;
}
int main() {
  int q;
  scanf("%d%d%s", &n, &q, s + 1);
  reverse(s + 1, s + n + 1);
  for (int i = 1; i <= n; ++i) append(s[i] - 'a'), poi[i] = np;
  get_edge();
  dfs(1);
  bui_st();
  while (q--) {
    ini2();
    printf("%lld\n", solve());
  }
}