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 = 400005, maxt = 1 << 19;
int sa[maxn], fst[maxn], sec[maxn], cnt[maxn], n;
char str[maxn];
int cmp(int a, int b, int i) {
  if (a + 2 * i > n || b + 2 * i > n) return 1;
  return sec[a] != sec[b] || sec[a + i] != sec[b + i];
}
void init_sa() {
  int p = 128;
  for (int i = 0; i < n; i++) ++cnt[fst[i] = str[i]];
  for (int i = 1; i < p; i++) cnt[i] += cnt[i - 1];
  for (int i = n - 1; i >= 0; i--) sa[--cnt[fst[i]]] = i;
  for (int i = 1; i < n; i <<= 1, ++p) {
    int q = 0;
    for (int j = n - i; j < n; j++) sec[q++] = j;
    for (int j = 0; j < n; j++)
      if (sa[j] >= i) sec[q++] = sa[j] - i;
    memset(cnt, 0, sizeof(cnt));
    for (int j = 0; j < n; j++) ++cnt[fst[sec[j]]];
    for (int j = 1; j < p; j++) cnt[j] += cnt[j - 1];
    for (int j = n - 1; j >= 0; j--) sa[--cnt[fst[sec[j]]]] = sec[j];
    memcpy(sec, fst, sizeof(sec));
    fst[sa[0]] = p = 0;
    for (int j = 1; j < n; j++) fst[sa[j]] = (p += cmp(sa[j - 1], sa[j], i));
  }
  for (int i = n; i > 0; i--) sa[i] = sa[i - 1];
  sa[0] = n;
}
int lcp[maxn], mn[maxt], tn;
void init_lcp() {
  for (int i = 0; i <= n; i++) fst[sa[i]] = i;
  int h = 0;
  for (int i = 0; i < n; i++) {
    int j = sa[fst[i] - 1];
    if (h > 0) --h;
    while (i + h < n && j + h < n && str[i + h] == str[j + h]) ++h;
    lcp[fst[i] - 1] = h;
  }
}
void update(int x, int w) {
  x += tn - 1;
  mn[x] = w;
  while (x > 0) {
    x = (x - 1) >> 1;
    mn[x] = min(mn[x * 2 + 1], mn[x * 2 + 2]);
  }
}
int query(int a, int b, int l = 0, int r = tn, int k = 0) {
  if (a <= l && b >= r) return mn[k];
  if (a >= r || b <= l) return INT_MAX;
  int mid = (l + r) >> 1;
  return min(query(a, b, l, mid, k * 2 + 1), query(a, b, mid, r, k * 2 + 2));
}
struct Query {
  int type, pla;
  bool operator<(const Query &q) const { return fst[pla] < fst[q.pla]; }
} qu[maxn];
pair<int, int> sta[maxn];
int main() {
  int m;
  scanf("%d%d", &n, &m);
  scanf("%s", str);
  init_sa(), init_lcp();
  memset(mn, 0x3f, sizeof(mn));
  for (tn = 1; tn < n; tn <<= 1)
    ;
  for (int i = 0; i < n; i++) update(i, lcp[i]);
  while (m--) {
    int na, nb;
    scanf("%d%d", &na, &nb);
    for (int i = 0; i < na; i++) {
      scanf("%d", &qu[i].pla);
      --qu[i].pla;
      qu[i].type = 1;
    }
    for (int i = na; i < na + nb; i++) {
      scanf("%d", &qu[i].pla);
      --qu[i].pla;
      qu[i].type = 0;
    }
    na += nb;
    sort(qu, qu + na);
    long long sum = 0, res = 0;
    int top = 0;
    for (int i = 0; i < na; i++) {
      if (i > 0) {
        int mn = query(fst[qu[i - 1].pla], fst[qu[i].pla]);
        if (mn < INT_MAX) {
          int cnt = 0;
          while (top > 0 && sta[top].first >= mn) {
            pair<int, int> p = sta[top--];
            sum -= (long long)p.second * p.first;
            cnt += p.second;
          }
          sta[++top] = pair<int, int>(mn, cnt);
          sum += (long long)mn * cnt;
        }
      }
      if (qu[i].type)
        res += sum;
      else
        sta[++top] = pair<int, int>(n - qu[i].pla, 1), sum += n - qu[i].pla;
    }
    reverse(qu, qu + na);
    sum = top = 0;
    for (int i = 0; i < na; i++) {
      if (i > 0) {
        int mn = query(fst[qu[i].pla], fst[qu[i - 1].pla]);
        if (mn < INT_MAX) {
          int cnt = 0;
          while (top > 0 && sta[top].first >= mn) {
            pair<int, int> p = sta[top--];
            sum -= (long long)p.second * p.first;
            cnt += p.second;
          }
          sta[++top] = pair<int, int>(mn, cnt);
          sum += (long long)mn * cnt;
        }
      }
      if (qu[i].type)
        res += sum;
      else
        sta[++top] = pair<int, int>(n - qu[i].pla, 1), sum += n - qu[i].pla;
    }
    printf("%lld\n", res);
  }
  return 0;
}