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;
char s[202020];
int x[202020], y[202020];
int sa[202020], h[202020];
int mi[20][202020];
int lg[202020];
int t[202020];
struct query {
  int pp, tp;
  bool operator<(const query &q) const {
    if (pp != q.pp) return x[pp] < x[q.pp];
    return tp < q.tp;
  }
} qu[404040];
int st[2][202020], stv[2][202020], siz[2][202020], sn[2], n;
int lcp(int a, int b) {
  if (a == b) return n - a;
  a = x[a];
  b = x[b];
  if (a > b) swap(a, b);
  return min(mi[lg[b - a]][a + 1], mi[lg[b - a]][b - (1 << lg[b - a]) + 1]);
}
int main() {
  int q, m = 300, an, bn, tt;
  long long nv[2], ans;
  scanf("%d%d", &n, &q);
  scanf("%s", s);
  memset(t, 0, sizeof(t));
  for (int i = 0; i < n; i++) t[x[i] = s[i]]++;
  for (int i = 1; i < m; i++) t[i] += t[i - 1];
  x[n] = y[n] = -1;
  for (int i = n - 1; i >= 0; i--) sa[--t[x[i]]] = i;
  int p = 0, k;
  for (int i = 1;; i <<= 1) {
    p = 0;
    for (int j = n - i; j < n; j++) y[p++] = j;
    for (int j = 0; j < n; j++)
      if (sa[j] >= i) y[p++] = sa[j] - i;
    for (int j = 0; j < m; j++) t[j] = 0;
    for (int j = 0; j < n; j++) t[x[j]]++;
    for (int j = 1; j < m; j++) t[j] += t[j - 1];
    for (int j = n - 1; j >= 0; j--) sa[--t[x[y[j]]]] = y[j];
    swap(x, y);
    x[sa[0]] = 0;
    m = 0;
    for (int j = 1; j < n; j++)
      if (y[sa[j]] == y[sa[j - 1]] && y[sa[j] + i] == y[sa[j - 1] + i])
        x[sa[j]] = m;
      else
        x[sa[j]] = ++m;
    m++;
    if (m == n) break;
  }
  k = 0;
  for (int i = 0; i < n; i++) {
    if (x[i] == 0) {
      h[0] = 0;
      k = 0;
      continue;
    }
    while (s[i + k] == s[sa[x[i] - 1] + k]) k++;
    h[x[i]] = k;
    if (k) k--;
  }
  int cnt = 0;
  lg[1] = 0;
  for (int i = 2; i <= n; i++) lg[i] = lg[i >> 1] + 1;
  memset(mi, 0, sizeof(mi));
  for (int i = 1; i < n; i++) mi[0][i] = h[i];
  for (int i = 1; i < 18; i++)
    for (int j = n - (1 << i); j >= 1; j--)
      mi[i][j] = min(mi[i - 1][j], mi[i - 1][j + (1 << i - 1)]);
  for (int i = 1; i <= q; i++) {
    scanf("%d%d", &an, &bn);
    cnt = 0;
    for (int j = 1; j <= an; j++) {
      ++cnt;
      scanf("%d", &qu[cnt].pp);
      qu[cnt].pp--;
      qu[cnt].tp = 0;
    }
    for (int j = 1; j <= bn; j++) {
      ++cnt;
      scanf("%d", &qu[cnt].pp);
      qu[cnt].pp--;
      qu[cnt].tp = 1;
    }
    sort(qu + 1, qu + cnt + 1);
    sn[0] = sn[1] = 0;
    nv[0] = nv[1] = 0;
    stv[0][0] = stv[1][0] = 0;
    ans = 0;
    for (int j = 1; j <= cnt; j++) {
      for (tt = 0; tt < 2; tt++) {
        while (sn[tt] && lcp(qu[j].pp, st[tt][sn[tt]]) <= stv[tt][sn[tt] - 1]) {
          nv[tt] += (long long)(stv[tt][sn[tt] - 1] - stv[tt][sn[tt]]) *
                    siz[tt][sn[tt]];
          siz[tt][sn[tt] - 1] += siz[tt][sn[tt]];
          sn[tt]--;
        }
        if (sn[tt]) {
          nv[tt] -= (long long)stv[tt][sn[tt]] * siz[tt][sn[tt]];
          stv[tt][sn[tt]] = lcp(qu[j].pp, st[tt][sn[tt]]);
          nv[tt] += (long long)stv[tt][sn[tt]] * siz[tt][sn[tt]];
        }
      }
      tt = qu[j].tp;
      st[tt][++sn[tt]] = qu[j].pp;
      stv[tt][sn[tt]] = 0;
      siz[tt][sn[tt]] = 1;
      ans += nv[tt ^ 1];
    }
    printf("%lld\n", ans);
  }
  return 0;
}