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 = 220000;
const int MAXS = 256;
const int MAXB = 20;
int n, q;
char s[MAXN];
int a[MAXN], b[MAXN], c[2 * MAXN];
bool d[2 * MAXN];
int sa[MAXN], rnk[2 * MAXN], rnk1[2 * MAXN];
int ct[MAXS], cnt[MAXN], tmp[MAXN];
int height[MAXN], lg[MAXN], h[MAXN][MAXB];
long long sum[2 * MAXN];
struct Node {
  int id, pos;
  Node(int i = 0, int p = 0) : id(i), pos(p) {}
} sta[2 * MAXN];
int top;
inline bool cmp(int a, int b) { return rnk[a] < rnk[b]; }
inline int query(int l, int r) {
  int k = lg[r - l + 1];
  return min(h[l][k], h[r - (1 << k) + 1][k]);
}
inline int lcp(int i, int j) {
  if (i == j) return n - i + 1;
  if (rnk[i] > rnk[j]) swap(i, j);
  return query(rnk[i] + 1, rnk[j]);
}
int main() {
  scanf("%d%d%s", &n, &q, s + 1);
  memset(ct, 0, sizeof ct);
  memset(rnk, 0, sizeof rnk);
  for (int i = 1; i <= n; i++) ct[s[i]] = 1;
  for (int i = 1; i < MAXS; i++) ct[i] += ct[i - 1];
  for (int i = 1; i <= n; i++) rnk[i] = ct[s[i]];
  for (int k = 0, p = 1; k != n; p <<= 1) {
    memset(cnt, 0, sizeof cnt);
    for (int i = 1; i <= n; i++) cnt[rnk[i + p]]++;
    for (int i = 1; i <= n; i++) cnt[i] += cnt[i - 1];
    for (int i = n; i >= 1; i--) tmp[cnt[rnk[i + p]]--] = i;
    memset(cnt, 0, sizeof cnt);
    for (int i = 1; i <= n; i++) cnt[rnk[i]]++;
    for (int i = 1; i <= n; i++) cnt[i] += cnt[i - 1];
    for (int i = n; i >= 1; i--) sa[cnt[rnk[tmp[i]]]--] = tmp[i];
    memcpy(rnk1, rnk, sizeof rnk1);
    rnk[sa[1]] = k = 1;
    for (int i = 2; i <= n; i++) {
      if (rnk1[sa[i]] != rnk1[sa[i - 1]] ||
          rnk1[sa[i] + p] != rnk1[sa[i - 1] + p])
        k++;
      rnk[sa[i]] = k;
    }
  }
  for (int i = 1, k = 0; i <= n; i++) {
    if (rnk[i] == 1) {
      height[rnk[i]] = k = 0;
      continue;
    }
    if (--k < 0) k = 0;
    while (s[i + k] == s[sa[rnk[i] - 1] + k]) k++;
    height[rnk[i]] = k;
  }
  lg[0] = 0;
  for (int i = 1; i <= n; i++) lg[i] = lg[i - 1] + (1 << lg[i - 1] + 1 == i);
  for (int i = 1; i <= n; i++) h[i][0] = height[i];
  for (int j = 1; 1 << j <= n; j++)
    for (int i = 1; i + (1 << j) - 1 <= n; i++)
      h[i][j] = min(h[i][j - 1], h[i + (1 << j - 1)][j - 1]);
  while (q--) {
    int k, l;
    scanf("%d%d", &k, &l);
    for (int i = 1; i <= k; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= l; i++) scanf("%d", &b[i]);
    sort(a + 1, a + k + 1, cmp);
    sort(b + 1, b + l + 1, cmp);
    for (int i = 1, j = 1, t = 1; i <= k || j <= l; t++) {
      if (i <= k && (j > l || rnk[a[i]] <= rnk[b[j]]))
        c[t] = a[i++], d[t] = 0;
      else if (j <= l && (i > k || rnk[a[i]] > rnk[b[j]]))
        c[t] = b[j++], d[t] = 1;
    }
    long long ans = 0;
    sum[0] = 0;
    top = 0;
    for (int i = 1; i <= k + l; i++) sum[i] = sum[i - 1] + d[i];
    for (int i = 1; i <= k + l; i++) {
      int t = i;
      while (top > 0 && lcp(c[i], c[i - 1]) <
                            lcp(c[sta[top - 1].id], c[sta[top - 1].id - 1])) {
        t = min(t, sta[--top].pos);
        ans += (1ll * (sum[sta[top].id - 1] - sum[sta[top].pos - 2]) *
                    ((i - sta[top].id) - (sum[i - 1] - sum[sta[top].id - 1])) +
                1ll *
                    ((sta[top].id - sta[top].pos + 1) -
                     (sum[sta[top].id - 1] - sum[sta[top].pos - 2])) *
                    (sum[i - 1] - sum[sta[top].id - 1])) *
               lcp(c[sta[top].id], c[sta[top].id - 1]);
      }
      sta[top++] = Node(i, t);
    }
    while (top > 0) {
      top--;
      ans += (1ll * (sum[sta[top].id - 1] - sum[sta[top].pos - 2]) *
                  ((k + l - sta[top].id + 1) -
                   (sum[k + l] - sum[sta[top].id - 1])) +
              1ll *
                  ((sta[top].id - sta[top].pos + 1) -
                   (sum[sta[top].id - 1] - sum[sta[top].pos - 2])) *
                  (sum[k + l] - sum[sta[top].id - 1])) *
             lcp(c[sta[top].id], c[sta[top].id - 1]);
    }
    printf("%lld\n", ans);
  }
  return 0;
}