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 = 210005;
int n;
int wa[maxn], wb[maxn], wv[maxn], brac[maxn];
int cmp(int *r, int a, int b, int l) {
  return r[a] == r[b] && r[a + l] == r[b + l];
}
void da(char *r, int *sa, int n, int m) {
  int i, j, p, *x = wa, *y = wb, *t;
  for (i = 0; i < m; i++) brac[i] = 0;
  for (i = 0; i < n; i++) brac[x[i] = r[i]]++;
  for (i = 1; i < m; i++) brac[i] += brac[i - 1];
  for (i = n - 1; i >= 0; i--) sa[--brac[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++) brac[i] = 0;
    for (i = 0; i < n; i++) brac[wv[i]]++;
    for (i = 1; i < m; i++) brac[i] += brac[i - 1];
    for (i = n - 1; i >= 0; i--) sa[--brac[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++;
  }
  return;
}
int rk[maxn], height[maxn];
void calheight(char *r, int *sa, int n) {
  int i, j, k = 0;
  for (i = 1; i <= n; i++) rk[sa[i]] = i;
  for (i = 0; i < n; height[rk[i++]] = k)
    for (k ? k-- : 0, j = sa[rk[i] - 1]; r[i + k] == r[j + k]; k++)
      ;
  return;
}
int RMQ[maxn];
int mm[maxn];
int best[20][maxn];
void initRMQ(int n) {
  int i, j, a, b;
  for (int i = 1; i <= n; i++) {
    RMQ[i] = height[i];
  }
  for (mm[0] = -1, i = 1; i <= n; i++)
    mm[i] = ((i & (i - 1)) == 0) ? mm[i - 1] + 1 : mm[i - 1];
  for (i = 1; i <= n; i++) best[0][i] = i;
  for (i = 1; i <= mm[n]; i++)
    for (j = 1; j <= n + 1 - (1 << i); j++) {
      a = best[i - 1][j];
      b = best[i - 1][j + (1 << (i - 1))];
      if (RMQ[a] < RMQ[b])
        best[i][j] = a;
      else
        best[i][j] = b;
    }
  return;
}
int askRMQ(int a, int b) {
  int t;
  t = mm[b - a + 1];
  b -= (1 << t) - 1;
  a = best[t][a];
  b = best[t][b];
  return RMQ[a] < RMQ[b] ? a : b;
}
int lcp(int a, int b) {
  if (a == b) {
    return n - a;
  }
  int t;
  a = rk[a];
  b = rk[b];
  if (a > b) {
    t = a;
    a = b;
    b = t;
  }
  return (height[askRMQ(a + 1, b)]);
}
char str[maxn];
int sa[maxn];
bool cmp1(pair<int, bool> &a, pair<int, bool> &b) {
  if (a.first != b.first) {
    return rk[a.first] < rk[b.first];
  }
  if (a.second) {
    return true;
  }
  return false;
}
long long cnts[maxn];
int main() {
  int len, q;
  scanf("%d%d", &len, &q);
  n = len;
  scanf("%s", str);
  da(str, sa, len + 1, 300);
  calheight(str, sa, len);
  initRMQ(len);
  while (q--) {
    stack<pair<int, int> > sta;
    int k, l;
    scanf("%d%d", &k, &l);
    vector<pair<int, bool> > arr;
    for (int i = 0; i < k; i++) {
      int a;
      scanf("%d", &a);
      a--;
      arr.push_back({a, true});
    }
    for (int i = 0; i < l; i++) {
      int a;
      scanf("%d", &a);
      a--;
      arr.push_back({a, false});
    }
    sort(arr.begin(), arr.end(), cmp1);
    int sz = arr.size();
    long long ans = 0;
    for (int k = 0; k < 2; k++) {
      long long sum = 0;
      map<int, int> mp;
      for (int i = 0; i < sz; i++) {
        auto pr = arr[i];
        if (k == pr.second) {
          ans += sum;
        } else {
          mp[n - pr.first]++;
          sum += n - pr.first;
        }
        if (i < sz - 1) {
          int tmp = lcp(arr[i].first, arr[i + 1].first);
          while (!mp.empty()) {
            auto it = --(mp.end());
            if (tmp >= it->first) {
              break;
            }
            sum -= 1ll * (it->first - tmp) * it->second;
            mp[tmp] += it->second;
            mp.erase(it);
          }
        }
      }
    }
    printf("%I64d\n", ans);
  }
  return 0;
}