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 N = 4e5 + 10;
int c[N], sa[N], x[N], y[N], rk[N], hei[N], he[N], n, m, cc[N << 1], st[N][20],
    a[N], h[N], T, b[N];
long long dp[N];
char s[N];
bool cmp(int x, int y) { return rk[x] < rk[y]; }
void get_sa() {
  for (int i = 1; i <= n; i++) ++c[x[i] = s[i]];
  for (int i = 2; i <= m; i++) c[i] += c[i - 1];
  for (int i = n; i >= 1; i--) sa[c[x[i]]--] = i;
  for (int k = 1; k <= n; k <<= 1) {
    int num = 0;
    for (int i = n - k + 1; i <= n; i++) y[++num] = i;
    for (int i = 1; i <= n; i++)
      if (sa[i] > k) y[++num] = sa[i] - k;
    for (int i = 1; i <= m; i++) c[i] = 0;
    for (int i = 1; i <= n; i++) ++c[x[i]];
    for (int i = 2; i <= m; i++) c[i] += c[i - 1];
    for (int i = n; i >= 1; i--) sa[c[x[y[i]]]--] = y[i], y[i] = 0;
    swap(x, y);
    x[sa[1]] = 1;
    num = 1;
    for (int i = 2; i <= n; i++)
      x[sa[i]] = (y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k])
                     ? num
                     : ++num;
    if (num == n) break;
    m = num;
  }
}
void get_h() {
  memset(hei, 0, sizeof(hei));
  for (int i = 1; i <= n; i++) rk[sa[i]] = i;
  int k = 0;
  for (int i = 1; i <= n; i++) {
    if (rk[i] == 1) continue;
    if (k) k--;
    int j = sa[rk[i] - 1];
    while (i + k <= n && j + k <= n && s[i + k] == s[j + k]) k++;
    hei[rk[i]] = k;
    st[rk[i]][0] = k;
  }
}
int lcp(int l, int r, int len) {
  if (l == r) return len - l + 1;
  int x = rk[l], y = rk[r];
  if (x > y) swap(x, y);
  x++;
  int k = log2(y - x + 1);
  return min(st[x][k], st[y - (1 << k) + 1][k]);
}
long long work(int len) {
  sort(a + 1, a + len + 1, cmp);
  for (int i = 2; i <= len; i++) he[i] = lcp(a[i], a[i - 1], n);
  int top = 0;
  long long ans = 0;
  for (int i = 1; i <= len; i++) {
    while (top > 0 && he[h[top]] >= he[i]) top--;
    int p = h[top];
    dp[i] = dp[p] + (long long)(i - p) * he[i];
    ans += dp[i];
    h[++top] = i;
  }
  for (int i = 1; i <= len; i++) dp[i] = 0, h[i] = 0, he[i] = 0;
  return ans;
}
int main() {
  scanf("%d%d", &n, &T);
  scanf("%s", s + 1);
  m = 122;
  get_sa();
  get_h();
  for (int j = 1; j <= 18; j++)
    for (int i = 1; i + (1 << j) - 1 <= n; i++)
      st[i][j] = min(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
  while (T--) {
    int la, lb;
    scanf("%d%d", &la, &lb);
    long long ans = 0;
    int cnt = 0;
    for (int i = 1; i <= la; i++) scanf("%d", &a[i]), cc[++cnt] = a[i];
    for (int i = 1; i <= lb; i++) scanf("%d", &b[i]), cc[++cnt] = b[i];
    ans -= work(la);
    for (int i = 1; i <= lb; i++) a[i] = b[i];
    ans -= work(lb);
    for (int i = 1; i <= la + lb; i++) a[i] = cc[i];
    ans += work(la + lb);
    printf("%I64d\n", ans);
  }
  return 0;
}