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, m, Q, a[200010], b[200010], q[200010], p[200010];
char s[200010];
long long ans;
int cnt[200010], x[200010], y[200010], t[200010], sa[200010], g[200010][20],
    rk[200010], height[200010];
struct info {
  int x, tp, k;
  bool operator<(const info &p) const {
    if (k) return rk[x] < rk[p.x] || (x == p.x && tp > p.tp);
    return rk[x] > rk[p.x] || (x == p.x && tp < p.tp);
  }
} h[200010 * 2];
bool cmp(int *g, int a, int b, int l) {
  return g[a] == g[b] && g[a + l] == g[b + l];
}
void get_sa() {
  m = 128;
  for (int i = 1; i <= n; i++) cnt[x[i] = s[i]]++;
  for (int i = 1; i <= m; i++) cnt[i] += cnt[i - 1];
  for (int i = n; i; i--) sa[cnt[x[i]]--] = i;
  for (int j = 1, tot = 0; tot <= n; m = tot, j <<= 1) {
    tot = 0;
    for (int i = n - j + 1; i <= n; i++) y[++tot] = i;
    for (int i = 1; i <= n; i++)
      if (sa[i] > j) y[++tot] = sa[i] - j;
    for (int i = 1; i <= n; i++) t[i] = x[y[i]];
    memset(cnt, 0, sizeof(cnt));
    for (int i = 1; i <= n; i++) cnt[t[i]]++;
    for (int i = 1; i <= m; i++) cnt[i] += cnt[i - 1];
    for (int i = n; i; i--) sa[cnt[t[i]]--] = y[i];
    tot = 2;
    swap(x, y);
    x[sa[1]] = 1;
    for (int i = 2; i <= n; i++)
      x[sa[i]] = cmp(y, sa[i], sa[i - 1], j) ? tot - 1 : tot++;
  }
}
void get_height() {
  for (int i = 1; i <= n; i++) rk[sa[i]] = i;
  for (int i = 1, j, k = 0; i <= n; height[rk[i++]] = k)
    for (k = k ? k - 1 : 0, j = sa[rk[i] - 1]; s[i + k] == s[j + k]; k++)
      ;
}
void get_rmq() {
  memset(g, 127, sizeof(g));
  for (int i = 1; i <= n; i++) g[i][0] = height[i];
  for (int j = 1; (1 << j) <= n; j++)
    for (int i = 1; i <= n - (1 << j) + 1; i++)
      g[i][j] = min(g[i][j - 1], g[i + (1 << j - 1)][j - 1]);
}
int lcp(int x, int y) {
  if (x == y) return n - x + 1;
  x = rk[x];
  y = rk[y];
  if (x > y) swap(x, y);
  x++;
  int p = log2(y - x + 1);
  return min(g[x][p], g[y - (1 << p) + 1][p]);
}
long long solve(int l1, int l2, int tp) {
  long long sum = 0, ans = 0;
  int tot = 0, top = 0;
  for (int i = 1; i <= l1; i++) h[++tot] = (info){a[i], 0, tp};
  for (int i = 1; i <= l2; i++) h[++tot] = (info){b[i], 1, tp};
  sort(h + 1, h + tot + 1);
  for (int i = 2, j = h[1].tp; i <= tot; i++) {
    int tmp = lcp(h[i - 1].x, h[i].x);
    while (top && tmp <= q[top]) sum -= (p[top] - p[top - 1]) * q[top], top--;
    sum += tmp * (j - p[top]);
    q[++top] = tmp;
    p[top] = j;
    if (!h[i].tp) ans += sum;
    j += h[i].tp;
  }
  return ans;
}
int main() {
  int l1, l2;
  scanf("%d%d %s", &n, &Q, s + 1);
  get_sa();
  get_height();
  get_rmq();
  while (Q--) {
    scanf("%d%d", &l1, &l2);
    ans = 0;
    for (int i = 1; i <= l1; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= l2; i++) scanf("%d", &b[i]);
    b[0] = 1;
    b[l2 + 1] = n;
    ans += solve(l1, l2, 1);
    reverse(a, a + l1 + 2);
    reverse(b, b + l2 + 2);
    ans += solve(l1, l2, 0);
    printf("%I64d\n", ans);
  }
  return 0;
}