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;
int n, q, mat[400005], h[400005], len[400005], tot, fa[400005];
char s[400005];
struct edge {
  int to, nxt;
} e[400005];
void add(int x, int y) {
  e[++tot].nxt = h[x];
  h[x] = tot;
  e[tot].to = y;
}
struct SAM {
  int idx, pre, ch[400005][26], T;
  SAM() { fa[0] = ~0; }
  void extend(int x) {
    int cur = ++idx, p = pre;
    len[cur] = len[p] + 1;
    while (~p && !ch[p][x]) ch[p][x] = cur, p = fa[p];
    if (-1 == p)
      fa[cur] = 0;
    else {
      int q = ch[p][x];
      if (len[p] + 1 == len[q])
        fa[cur] = q;
      else {
        int now = ++idx;
        len[now] = len[p] + 1;
        for (int i = 0; i <= 25; i++) ch[now][i] = ch[q][i];
        fa[now] = fa[q];
        fa[q] = fa[cur] = now;
        while (~p && ch[p][x] == q) ch[p][x] = now, p = fa[p];
      }
    }
    mat[++T] = pre = cur;
  }
  void init() {
    for (int i = 1; i <= idx; i++) add(fa[i], i);
  }
} w;
int dfn[400005], idx, top[400005], sz[400005], son[400005], t, d[400005];
void dfs(int x) {
  sz[x] = 1;
  if (x) d[x] = d[fa[x]] + 1;
  for (int i = h[x]; i; i = e[i].nxt) {
    dfs(e[i].to);
    sz[x] += sz[e[i].to];
    if (!son[x] || sz[son[x]] < sz[e[i].to]) son[x] = e[i].to;
  }
}
void dfs2(int x, int tp) {
  dfn[x] = ++idx;
  top[x] = tp;
  if (!son[x]) return;
  dfs2(son[x], tp);
  for (int i = h[x]; i; i = e[i].nxt)
    if (e[i].to != son[x]) dfs2(e[i].to, e[i].to);
}
int lca(int x, int y) {
  while (top[x] != top[y]) {
    if (d[top[x]] < d[top[y]]) swap(x, y);
    x = fa[top[x]];
  }
  if (d[x] > d[y]) swap(x, y);
  return x;
}
int u[400005], v[400005], ll[400005], rr[400005];
pair<int, int> c[400005];
vector<int> a[400005];
int sta[400005], tp;
void ins(int x) {
  int z = lca(sta[tp], x);
  if (z == sta[tp])
    sta[++tp] = x;
  else {
    while (dfn[sta[tp]] > dfn[z]) {
      if (dfn[sta[tp - 1]] >= dfn[z])
        a[sta[tp - 1]].push_back(sta[tp]), tp--;
      else
        a[z].push_back(sta[tp]), sta[tp] = z;
    }
    sta[++tp] = x;
  }
}
long long ans;
void dfs3(int x, int f) {
  for (int i = 0; i < (int)a[x].size(); i++) {
    dfs3(a[x][i], x);
    ll[x] += ll[a[x][i]];
    rr[x] += rr[a[x][i]];
  }
  if (x) ans += 1LL * ll[x] * rr[x] * (len[x] - len[f]);
}
void clear(int x) {
  ll[x] = rr[x] = 0;
  for (int i = 0; i < (int)a[x].size(); i++) clear(a[x][i]);
  a[x].clear();
}
int main() {
  scanf("%d%d", &n, &q);
  scanf("%s", s + 1);
  for (int i = n; i >= 1; i--) w.extend(s[i] - 'a');
  w.init();
  dfs(0);
  dfs2(0, 0);
  while (q--) {
    int k, l, t = 0;
    scanf("%d%d", &k, &l);
    for (int i = 1; i <= k; i++)
      scanf("%d", &u[i]), u[i] = n + 1 - u[i],
                          c[++t] = make_pair(dfn[mat[u[i]]], mat[u[i]]);
    for (int i = 1; i <= l; i++)
      scanf("%d", &v[i]), v[i] = n + 1 - v[i],
                          c[++t] = make_pair(dfn[mat[v[i]]], mat[v[i]]);
    sort(c + 1, c + t + 1);
    sta[1] = 0;
    tp = 1;
    for (int i = 1; i <= t; i++) {
      if (c[i] != c[i - 1]) ins(c[i].second);
    }
    while (tp > 1) a[sta[tp - 1]].push_back(sta[tp]), tp--;
    for (int i = 1; i <= k; i++) ll[mat[u[i]]]++;
    for (int i = 1; i <= l; i++) rr[mat[v[i]]]++;
    ans = 0;
    dfs3(0, 0);
    printf("%lld\n", ans);
    clear(0);
  }
  return 0;
}