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 mod = 1e9 + 7;
const int N = 200010;
char ss[N];
int dp[N][20];
struct node {
  int lcp, id, t;
} stk[N];
std::vector<pair<int, int> > v;
int sa[N], Rank[N], h[N], n;
int xx[N], yy[N], c[N];
char *s;
bool cmp(int *s, int x, int y, int k) {
  return (s[x] == s[y]) && (s[x + k] == s[y + k]);
}
void ins(char *str) {
  s = str;
  n = strlen(s) + 1;
}
void DA() {
  memset(c, 0, sizeof(c));
  int *x = xx, *y = yy, m = 130, *t, i;
  for (i = 0; i < n; i++) x[i] = s[i];
  for (i = 0; i < n; i++) c[x[i]]++;
  for (i = 1; i < m; i++) c[i] += c[i - 1];
  for (i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;
  for (int k = 1, tot = 0; tot < n; k <<= 1, m = tot) {
    memset(c, 0, sizeof(c));
    for (i = 0; i < n; i++) c[x[i]]++;
    for (i = 1; i < m; i++) c[i] += c[i - 1];
    for (i = n - k, tot = 0; i < n; i++) y[tot++] = i;
    for (i = 0; i < n; i++)
      if (sa[i] >= k) y[tot++] = sa[i] - k;
    for (i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
    for (i = tot = 1, t = x, x = y, y = t, x[sa[0]] = 0; i < n; i++)
      x[sa[i]] = cmp(y, sa[i - 1], sa[i], k) ? tot - 1 : tot++;
  }
}
void cal_height() {
  int i, j, k = 0;
  for (i = 1; i < n; i++) Rank[sa[i]] = i;
  for (i = 0; i < n - 1; h[Rank[i++]] = k)
    for (k ? k-- : 0, j = sa[Rank[i] - 1]; s[i + k] == s[j + k]; k++)
      ;
  int m = floor(log(n + 0.0) / log(2.0));
  for (int i = 1; i <= n; i++) dp[i][0] = h[i];
  for (int i = 1; i <= m; i++)
    for (int j = n; j; j--) {
      dp[j][i] = dp[j][i - 1];
      if (j + (1 << (i - 1)) <= n)
        dp[j][i] = min(dp[j][i], dp[j + (1 << (i - 1))][i - 1]);
    }
}
inline int lcp(int l, int r) {
  int m = 31 - __builtin_clz(r - l + 1);
  return min(dp[l][m], dp[r - (1 << m) + 1][m]);
}
inline bool cmpp(pair<int, int> a, pair<int, int> b) {
  return a.first == b.first ? a.second > b.second : a.first < b.first;
}
unordered_map<int, bool> mp;
int main(int argc, char const *argv[]) {
  int n, q;
  scanf("%d%d", &n, &q);
  scanf("%s", ss);
  ins(ss);
  DA();
  cal_height();
  while (q--) {
    long long ans = 0;
    v.clear();
    mp.clear();
    int k, l;
    scanf("%d%d", &k, &l);
    for (int i = 1; i <= k; i++) {
      int x;
      scanf("%d", &x);
      mp[x] = 1;
      x--;
      v.push_back({Rank[x] + 1, 1});
    }
    for (int i = 1; i <= l; i++) {
      int x;
      scanf("%d", &x);
      if (mp.count(x)) ans += n - x + 1;
      x--;
      v.push_back({Rank[x] + 1, 0});
    }
    sort(v.begin(), v.end());
    for (int t = 0; t < 2; t++) {
      long long res = 0;
      int top = 0;
      for (auto i : v) {
        int x = i.first, y = i.second;
        node now = {h[x], x, y ^ t};
        if (!(y ^ t)) {
          if (!top || x == stk[top].id) continue;
          if (x - 1 >= stk[top].id + 1) {
            int LCP = lcp(stk[top].id + 1, x - 1), cnt = 0;
            while (top && LCP <= stk[top].lcp) {
              res -= (long long)stk[top].t * (stk[top].lcp - LCP);
              cnt += stk[top--].t;
            }
            if (cnt && LCP) stk[++top] = {LCP, x - 1, cnt};
          }
          ans += res;
        } else {
          int LCP = now.lcp, cnt = 0;
          if (top)
            if (stk[top].id + 1 <= x) LCP = min(LCP, lcp(stk[top].id + 1, x));
          res += now.lcp;
          while (top && LCP <= stk[top].lcp) {
            res -= (long long)stk[top].t * (stk[top].lcp - LCP);
            cnt += stk[top--].t;
          }
          if (cnt && LCP) stk[++top] = {LCP, x, cnt};
          stk[++top] = now;
        }
      }
      if (!t) sort(v.begin(), v.end(), cmpp);
    }
    printf("%lld\n", ans);
  }
  return 0;
}