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 ssize;
int wa[211111], wb[211111], gap, wss[211111], wv[211111], pos[211111],
    s[211111], sa[211111];
int cmp(int *r, int a, int b, int l) {
  return r[a] == r[b] && r[a + l] == r[b + l];
}
void da(int *r, int *sa, int n, int m) {
  int i, j, p, *x = wa, *y = wb, *t;
  for (i = 0; i < m; i++) wss[i] = 0;
  for (i = 0; i < n; i++) wss[x[i] = r[i]]++;
  for (i = 1; i < m; i++) wss[i] += wss[i - 1];
  for (i = n - 1; i >= 0; i--) sa[--wss[x[i]]] = i;
  for (j = 1, p = 1; p < n; j *= 2, m = p) {
    for (p = 0, i = n - j; i < n; i++) y[p++] = i;
    for (i = 0; i < n; i++)
      if (sa[i] >= j) y[p++] = sa[i] - j;
    for (i = 0; i < n; i++) wv[i] = x[y[i]];
    for (i = 0; i < m; i++) wss[i] = 0;
    for (i = 0; i < n; i++) wss[wv[i]]++;
    for (i = 1; i < m; i++) wss[i] += wss[i - 1];
    for (i = n - 1; i >= 0; i--) sa[--wss[wv[i]]] = y[i];
    for (t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; i++)
      x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;
  }
}
void geth(int height[], int n, int sa[]) {
  for (int i = 0; i < n; i++) pos[sa[i]] = i;
  for (int i = 0, h = height[0] = 0; i < n - 1; i++) {
    int j = sa[pos[i] - 1];
    while (i + h < n && j + h < n && s[i + h] == s[j + h]) h++;
    if (height[pos[i]] = h) h--;
  }
}
char tmps[211111];
int height[211111];
int LOG[211111];
int f[18][211111];
void getlist() {
  LOG[0] = -1;
  for (int i = 1; i <= ssize; i++) LOG[i] = LOG[i / 2] + 1;
}
void create() {
  for (int j = 1; j <= ssize; j++) f[0][j] = height[j];
  for (int i = 1; i <= 17; i++)
    for (int j = 1; j + (1 << i) - 1 <= ssize; j++)
      f[i][j] = min(f[i - 1][j], f[i - 1][j + (1 << (i - 1))]);
}
int query(int qleft, int qright) {
  int nLog = LOG[qright - qleft + 1];
  return min(f[nLog][qleft], f[nLog][qright - (1 << nLog) + 1]);
}
vector<int> va, vb;
long long res = 0;
pair<int, int> st[211111];
int top;
void addit(int bef, int nowpos, long long &sumnow) {
  int co = 0, k = query(bef + 1, nowpos);
  while (top && st[top - 1].first >= k)
    co += st[top - 1].second,
        sumnow -= st[top - 1].first * 1LL * st[top - 1].second, top--;
  if (co) sumnow += co * 1LL * k, st[top++] = pair<int, int>(k, co);
}
void doit(vector<int> &va, vector<int> &vb) {
  int nowb = 0, bef = 0;
  top = 0;
  long long sumnow = 0;
  for (int k1 : va) {
    while (nowb != vb.size() && vb[nowb] <= k1) {
      addit(bef, vb[nowb], sumnow);
      int x = ssize - sa[vb[nowb]];
      st[top++] = {x, 1}, sumnow += x;
      bef = vb[nowb++];
    }
    if (k1 != bef) addit(bef, k1, sumnow), bef = k1;
    res += sumnow;
  }
}
int main() {
  int q, k1, k2, a, b;
  scanf("%d%d%s", &ssize, &q, tmps);
  for (int i = 0; i < ssize; i++) s[i] = tmps[i];
  ssize++;
  da(s, sa, ssize, 256);
  geth(height, ssize, sa);
  ssize--;
  getlist();
  create();
  while (q--) {
    scanf("%d%d", &k1, &k2);
    va.clear(), vb.clear();
    for (int i = 1; i < k1 + 1; i++) scanf("%d", &a), va.push_back(pos[a - 1]);
    for (int i = 1; i < k2 + 1; i++) scanf("%d", &a), vb.push_back(pos[a - 1]);
    sort(va.begin(), va.end());
    sort(vb.begin(), vb.end());
    res = 0;
    int nowb = 0;
    doit(va, vb), doit(vb, va);
    for (int k1 : va) {
      while (nowb != vb.size() && vb[nowb] < k1) nowb++;
      while (nowb != vb.size() && vb[nowb] == k1)
        res -= ssize - sa[vb[nowb]], nowb++;
    }
    printf("%I64d\n", res);
  }
}