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 Maxn = 200010;
const int inf = 2147483647;
int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
  return x * f;
}
char s[Maxn];
int n, q, lst = 1, tot = 1, mx[Maxn << 1], son[Maxn << 1][26], par[Maxn << 1];
int dfn[Maxn << 1], DFN = 0, dep[Maxn << 1], fa[Maxn << 1][19], to[Maxn];
void extend(int x) {
  int p = lst, np = ++tot;
  mx[np] = mx[p] + 1;
  while (p && !son[p][x]) son[p][x] = np, p = par[p];
  if (!p)
    par[np] = 1;
  else {
    int q = son[p][x];
    if (mx[p] + 1 == mx[q])
      par[np] = q;
    else {
      int nq = ++tot;
      mx[nq] = mx[p] + 1;
      for (int i = 0; i < 26; i++) son[nq][i] = son[q][i];
      par[nq] = par[q];
      par[q] = par[np] = nq;
      while (son[p][x] == q) son[p][x] = nq, p = par[p];
    }
  }
  lst = np;
}
struct Edge {
  int y, next;
} e[Maxn << 1], E[Maxn << 1];
int last[Maxn << 1], len = 0;
void ins(int x, int y) {
  int t = ++len;
  e[t].y = y;
  e[t].next = last[x];
  last[x] = t;
}
int Last[Maxn << 1], Len;
void Ins(int x, int y) {
  int t = ++Len;
  E[t].y = y;
  E[t].next = Last[x];
  Last[x] = t;
}
void dfs(int x) {
  dfn[x] = ++DFN;
  for (int i = 1; (1 << i) <= dep[x]; i++) fa[x][i] = fa[fa[x][i - 1]][i - 1];
  for (int i = last[x]; i; i = e[i].next) {
    int y = e[i].y;
    fa[y][0] = x;
    dep[y] = dep[x] + 1;
    dfs(y);
  }
}
int LCA(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (int i = 18; i >= 0; i--)
    if ((1 << i) <= dep[x] - dep[y]) x = fa[x][i];
  if (x == y) return x;
  for (int i = 18; i >= 0; i--)
    if ((1 << i) <= dep[x] && fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
  return fa[x][0];
}
int k, l, a[Maxn], b[Maxn], cnt[2][Maxn << 1];
struct Node {
  int x, o;
} A[Maxn << 1];
bool cmp(Node a, Node b) { return dfn[a.x] < dfn[b.x]; }
int sta[Maxn << 1], top;
long long ans;
int tmp[Maxn << 1], lt;
void build() {
  sort(A + 1, A + 1 + k + l, cmp);
  sta[top = 1] = 1;
  Len = lt = 0;
  for (int i = 1; i <= k + l; i++) {
    cnt[A[i].o][A[i].x]++;
    if (i > 1 && A[i].x == A[i - 1].x) continue;
    int t = LCA(A[i].x, sta[top]);
    if (t != sta[top]) {
      while (top > 1 && dep[sta[top - 1]] >= dep[t])
        Ins(sta[top - 1], sta[top]), top--;
      if (sta[top] != t) Ins(t, sta[top]), sta[top] = t;
    }
    sta[++top] = A[i].x;
  }
  while (top > 1) Ins(sta[top - 1], sta[top]), top--;
}
void DFS(int x) {
  tmp[++lt] = x;
  ans += (long long)mx[x] * cnt[0][x] * cnt[1][x];
  for (int i = Last[x]; i; i = E[i].next) {
    int y = E[i].y;
    DFS(y);
    ans += (long long)mx[x] * cnt[0][y] * cnt[1][x];
    ans += (long long)mx[x] * cnt[0][x] * cnt[1][y];
    cnt[0][x] += cnt[0][y], cnt[1][x] += cnt[1][y];
  }
}
int main() {
  n = read(), q = read();
  scanf("%s", s + 1);
  for (int i = n; i; i--) extend(s[i] - 'a');
  for (int i = 2; i <= tot; i++) ins(par[i], i);
  dep[1] = 0;
  dfs(1);
  int p = 1;
  for (int i = n; i; i--) {
    int x = s[i] - 'a';
    p = son[p][x];
    to[i] = p;
  }
  while (q--) {
    k = read(), l = read();
    for (int i = 1; i <= k; i++) A[i].x = to[read()], A[i].o = 0;
    for (int i = 1; i <= l; i++) A[k + i].x = to[read()], A[k + i].o = 1;
    build();
    ans = 0;
    DFS(1);
    for (int i = 1; i <= lt; i++) {
      int x = tmp[i];
      cnt[0][x] = cnt[1][x] = Last[x] = 0;
    }
    printf("%lld\n", ans);
  }
}