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 read() {
  char ch = 0;
  int x = 0, flag = 1;
  while (!isdigit(ch)) {
    ch = getchar();
    if (ch == '-') flag = -1;
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  }
  return x * flag;
}
char ch[550000];
int root = 1, size = 1, last = 1;
struct node {
  int len, link, nxt[26];
} s[550000];
void build(int k) {
  int cur = ++size, p = last;
  s[cur].len = s[p].len + 1;
  while (p && !s[p].nxt[k]) {
    s[p].nxt[k] = cur;
    p = s[p].link;
  }
  last = cur;
  if (!p) {
    s[cur].link = root;
    return;
  }
  int q = s[p].nxt[k];
  if (s[q].len == s[p].len + 1)
    s[cur].link = q;
  else {
    int clone = ++size;
    s[clone] = s[q];
    s[clone].len = s[p].len + 1;
    while (p && s[p].nxt[k] == q) {
      s[p].nxt[k] = clone;
      p = s[p].link;
    }
    s[q].link = s[cur].link = clone;
  }
}
struct edge {
  int to, nxt;
} e[550000 * 2];
int num, head[550000];
inline void add(int x, int y) {
  e[++num] = (edge){y, head[x]};
  head[x] = num;
  e[++num] = (edge){x, head[y]};
  head[y] = num;
}
int times, dfn[550000], dep[550000], nxt[550000][22];
void table(int x, int fa) {
  dfn[x] = ++times;
  dep[x] = dep[fa] + 1;
  for (int i = head[x]; i != -1; i = e[i].nxt) {
    int to = e[i].to;
    if (to == fa) continue;
    table(to, x);
    nxt[to][0] = x;
  }
}
int lca(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (int i = 20; i >= 0; i--)
    if (dep[nxt[x][i]] >= dep[y]) x = nxt[x][i];
  if (x == y) return x;
  for (int i = 20; i >= 0; i--)
    if (nxt[x][i] != nxt[y][i]) x = nxt[x][i], y = nxt[y][i];
  return nxt[x][0];
}
bool cmp(int a, int b) { return dfn[a] < dfn[b]; }
long long ans;
bool flag1[550000], flag2[550000];
int top, p[550000], f[550000], g[550000], id[550000], st[550000];
void insert(int x) {
  if (!top) {
    st[++top] = x;
    return;
  }
  int LCA = lca(x, st[top]);
  while (top > 1 && dep[st[top - 1]] >= dep[LCA])
    add(st[top], st[top - 1]), top--;
  if (st[top] != LCA) add(st[top], LCA), st[top] = LCA;
  st[++top] = x;
}
void dfs(int x, int fa) {
  f[x] = g[x] = 0;
  if (flag1[x]) f[x] = 1;
  if (flag2[x]) g[x] = 1;
  ans += (long long)s[x].len * (long long)f[x] * (long long)g[x];
  for (int i = head[x]; i != -1; i = e[i].nxt) {
    int to = e[i].to;
    if (to == fa) continue;
    dfs(to, x);
    ans += (long long)s[x].len * ((long long)f[x] * (long long)g[to] +
                                  (long long)f[to] * (long long)g[x]);
    f[x] += f[to];
    g[x] += g[to];
  }
  head[x] = -1;
  flag1[x] = flag2[x] = false;
}
int main() {
  num = -1;
  memset(head, -1, sizeof(head));
  int n = read(), m = read();
  scanf("%s", ch);
  reverse(ch, ch + n);
  for (int i = 0; i < n; i++) build(ch[i] - 'a'), id[n - 1 - i] = last;
  for (int i = 2; i <= size; i++) add(s[i].link, i);
  table(1, 1);
  for (int k = 1; k <= 20; k++)
    for (int i = 1; i <= size; i++) nxt[i][k] = nxt[nxt[i][k - 1]][k - 1];
  num = -1;
  memset(head, -1, sizeof(head));
  for (int o = 1; o <= m; o++) {
    int k1 = read(), k2 = read(), t = 0;
    for (int i = 1; i <= k1; i++) p[++t] = id[read() - 1], flag1[p[t]] = true;
    for (int i = 1; i <= k2; i++) p[++t] = id[read() - 1], flag2[p[t]] = true;
    sort(p + 1, p + t + 1);
    t = unique(p + 1, p + t + 1) - p - 1;
    sort(p + 1, p + t + 1, cmp);
    top = 0;
    for (int i = 1; i <= t; i++) insert(p[i]);
    while (top > 1) add(st[top], st[top - 1]), top--;
    ans = 0;
    dfs(st[top], 0);
    num = -1;
    printf("%lld\n", ans);
  }
  return 0;
}