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 = 200050;
struct suffixArrary {
  int c[maxn], sa[maxn], t[maxn], x[maxn];
  int n, m, height[maxn], rank[maxn];
  void init(char *str, int _n) {
    n = _n, m = 255;
    x[n + 1] = t[n + 1] = str[n + 1] = 0;
    for (int i = 0; i <= m; ++i) c[i] = 0;
    for (int i = 1; i <= n; ++i) c[x[i] = str[i]]++;
    for (int i = 1; i <= m; ++i) c[i] += c[i - 1];
    for (int i = n; i >= 1; --i) sa[c[x[i]]--] = i;
    for (int l = 1; l <= n; l <<= 1) {
      int cnt = 0;
      for (int i = n - l + 1; i <= n; ++i) t[++cnt] = i;
      for (int i = 1; i <= n; ++i)
        if (sa[i] > l) t[++cnt] = sa[i] - l;
      for (int i = 0; i <= m; ++i) c[i] = 0;
      for (int i = 1; i <= n; ++i) c[x[i]]++;
      for (int i = 1; i <= m; ++i) c[i] += c[i - 1];
      for (int i = n; i >= 1; --i) sa[c[x[t[i]]]--] = t[i];
      m = 0, t[sa[1]] = ++m;
      for (int i = 2; i <= n; ++i) {
        if (cmp(sa[i], sa[i - 1], l))
          t[sa[i]] = m;
        else
          t[sa[i]] = ++m;
      }
      swap(t, x);
      if (m == n) break;
    }
    int h = 0;
    for (int i = 1; i <= n; ++i) rank[sa[i]] = i;
    for (int i = 1; i <= n; ++i) {
      if (h) --h;
      int j = sa[rank[i] - 1];
      while (str[i + h] == str[j + h]) ++h;
      height[rank[i]] = h;
    }
  }
  bool cmp(int u, int v, int l) { return x[u] == x[v] && x[u + l] == x[v + l]; }
} sa;
struct RMQ {
  int lg[maxn], f[maxn][20];
  RMQ() {
    lg[2] = 1;
    for (int i = 3; i < maxn; ++i) lg[i] = lg[i >> 1] + 1;
  }
  void init(int *arr, int n) {
    for (int j = 0; (1 << j) <= n; ++j) {
      for (int i = 1; i <= n; ++i) {
        if (j == 0)
          f[i][j] = arr[i];
        else
          f[i][j] = min(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
      }
    }
  }
  int Qmin(int x, int y) {
    int k = lg[y - x + 1];
    return min(f[x][k], f[y - (1 << k) + 1][k]);
  }
} st;
char str[maxn];
void deb(int *arr, int n) {
  for (int i = 1; i <= n; ++i) cout << arr[i] << " ";
  cout << endl;
}
int lcp(int x, int y) {
  if (x == y) return sa.n - x + 1;
  x = sa.rank[x], y = sa.rank[y];
  if (x > y) swap(x, y);
  return st.Qmin(x + 1, y);
}
int s[maxn];
long long solve(vector<int> &a, int k) {
  sort(a.begin(), a.end(),
       [](int x, int y) { return sa.rank[x] < sa.rank[y]; });
  vector<int> height(k);
  for (int i = 1; i < k; ++i) height[i] = lcp(a[i], a[i - 1]);
  long long ans = 0;
  int top = 0;
  for (int i = 0; i < k; ++i) {
    while (top && height[i] < height[s[top]]) {
      ans += (long long)height[s[top]] * (i - s[top]) * (s[top] - s[top - 1]),
          --top;
    }
    s[++top] = i;
  }
  while (top >= 1)
    ans += (long long)height[s[top]] * (k - s[top]) * (s[top] - s[top - 1]),
        --top;
  return ans;
}
int main() {
  int n, q;
  scanf("%d%d%s", &n, &q, str + 1);
  sa.init(str, n);
  st.init(sa.height, n);
  while (q--) {
    int k, l;
    scanf("%d%d", &k, &l);
    vector<int> a(k), b(l), c(k + l);
    for (int i = 0; i < k; ++i) scanf("%d", &a[i]), c[i] = a[i];
    for (int i = 0; i < l; ++i) scanf("%d", &b[i]), c[i + k] = b[i];
    printf("%lld\n", solve(c, l + k) - solve(a, k) - solve(b, l));
  }
  return 0;
}