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 long long read() {
  long long s = 0, f = 1;
  char ch;
  while (!isdigit(ch = getchar())) (ch == '-') && (f = -f);
  for (s = ch ^ 48; isdigit(ch = getchar());
       s = (s << 1) + (s << 3) + (ch ^ 48))
    ;
  return s * f;
}
const int N = 5e5 + 5;
int n, q_, sam, las, cnt, tot;
int len[N], t[N][26], link[N];
char s[N];
int fid[N], col[N][2], dep[N], dfn[N], head[N], f[N][21];
struct edge {
  int to, nxt;
} e[N];
vector<int> v[N];
void add(int x, int y) {
  e[++cnt].nxt = head[x];
  head[x] = cnt;
  e[cnt].to = y;
}
void insert(int c) {
  int cur = ++sam, p;
  len[cur] = len[las] + 1;
  for (p = las; p && !t[p][c]; p = link[p]) t[p][c] = cur;
  if (!p)
    link[cur] = 1;
  else {
    int x = t[p][c];
    if (len[x] == len[p] + 1)
      link[cur] = x;
    else {
      int cl = ++sam;
      len[cl] = len[p] + 1;
      memcpy(t[cl], t[x], sizeof(t[x]));
      link[cl] = link[x];
      link[x] = link[cur] = cl;
      while (p && t[p][c] == x) t[p][c] = cl, p = link[p];
    }
  }
  las = cur;
}
void get_tree(int x, int Fa) {
  dfn[x] = ++tot;
  f[x][0] = Fa;
  dep[x] = dep[Fa] + 1;
  for (int i = 1; i <= 20; i++) f[x][i] = f[f[x][i - 1]][i - 1];
  for (int i = head[x]; i; i = e[i].nxt) get_tree(e[i].to, x);
}
void Build_Suffix_Tree() {
  reverse(s + 1, s + n + 1);
  sam = las = 1;
  for (int i = 1; i <= n; i++) insert(s[i] - 'a'), fid[n - i + 1] = las;
  for (int i = 2; i <= sam; i++) add(link[i], i);
  get_tree(1, 0);
}
int LCA(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (int i = 20; i >= 0; i--)
    if (dep[x] - dep[y] >= (1 << i)) x = f[x][i];
  if (x == y) return x;
  for (int i = 20; i >= 0; i--)
    if (f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
  return f[x][0];
}
int top, num, tmp[N], sta[N], g[N][2];
long long Ans;
void insert_xs(int x) {
  int lca = LCA(x, sta[top]);
  while (top > 1 && dfn[lca] <= dfn[sta[top - 1]])
    v[sta[top - 1]].push_back(sta[top]), top--;
  if (lca != sta[top]) v[lca].push_back(sta[top]), sta[top] = lca;
  sta[++top] = x;
}
int cmp(int x, int y) { return dfn[x] < dfn[y]; }
void Build_xs() {
  sort(tmp + 1, tmp + num + 1, cmp);
  sta[top = 1] = 1;
  for (int i = 1; i <= num; i++) insert_xs(tmp[i]);
  while (top > 1) v[sta[top - 1]].push_back(sta[top]), top--;
}
void DP(int x) {
  if (col[x][0]) g[x][0] = 1;
  if (col[x][1]) g[x][1] = 1;
  if (col[x][0] && col[x][1]) Ans += len[x];
  for (int i = 0; i < (int)v[x].size(); i++) {
    int y = v[x][i];
    DP(y);
    Ans += (1ll * g[x][0] * g[y][1] + 1ll * g[x][1] * g[y][0]) * len[x];
    g[x][0] += g[y][0];
    g[x][1] += g[y][1];
  }
}
void Clear(int x) {
  col[x][0] = col[x][1] = 0;
  g[x][0] = g[x][1] = 0;
  for (int i = 0; i < (int)v[x].size(); i++) Clear(v[x][i]);
  v[x].clear();
}
void Work(int X, int Y) {
  num = 0;
  for (int i = 1, x; i <= X; i++) tmp[++num] = x = fid[read()], col[x][0] = 1;
  for (int i = 1, x; i <= Y; i++) tmp[++num] = x = fid[read()], col[x][1] = 1;
  sort(tmp + 1, tmp + num + 1);
  num = unique(tmp + 1, tmp + num + 1) - tmp - 1;
  Build_xs();
  DP(1);
  printf("%lld\n", Ans);
  Ans = 0;
  Clear(1);
}
int main() {
  n = read();
  q_ = read();
  scanf("%s", s + 1);
  Build_Suffix_Tree();
  for (int i = 1, X, Y; i <= q_; i++) X = read(), Y = read(), Work(X, Y);
  return 0;
}