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>
inline int read() {
  int res = 0;
  bool bo = 0;
  char c;
  while (((c = getchar()) < '0' || c > '9') && c != '-')
    ;
  if (c == '-')
    bo = 1;
  else
    res = c - 48;
  while ((c = getchar()) >= '0' && c <= '9')
    res = (res << 3) + (res << 1) + (c - 48);
  return bo ? ~res + 1 : res;
}
template <class T>
inline void Swap(T &a, T &b) {
  T t = a;
  a = b;
  b = t;
}
template <class T>
inline T Min(const T &a, const T &b) {
  return a < b ? a : b;
}
const int N = 2e5 + 5, LogN = 20;
int n, q, k, l, a[N], b[N], sa[N], rank[N], height[N], Log[N], RMQ[N][LogN],
    w[N], u[N], v[N], cnt[N], val[N], top;
char s[N];
bool vis[N];
void init() {
  int m = 26, *x = rank, *y = height;
  for (int i = 1; i <= n; i++) w[x[i] = s[i] - 'a' + 1]++;
  for (int i = 2; i <= m; i++) w[i] += w[i - 1];
  for (int i = 1; i <= n; i++) sa[w[x[i]]--] = i;
  for (int k = 1; k < n; k <<= 1, Swap(x, y)) {
    int tt = 0;
    for (int i = n - k + 1; i <= n; i++) y[++tt] = i;
    for (int i = 1; i <= n; i++)
      if (sa[i] > k) y[++tt] = sa[i] - k;
    memset(w, 0, sizeof(w));
    for (int i = 1; i <= n; i++) w[x[i]]++;
    for (int i = 2; i <= m; i++) w[i] += w[i - 1];
    for (int i = n; i >= 1; i--) sa[w[x[y[i]]]--] = y[i];
    m = 0;
    for (int i = 1; i <= n; i++) {
      int u = sa[i], v = sa[i - 1];
      y[u] = x[u] != x[v] || x[u + k] != x[v + k] ? ++m : m;
    }
    if (m == n) break;
  }
  for (int i = 1; i <= n; i++) rank[sa[i]] = i;
  for (int i = 1, k = 0; i <= n; i++) {
    if (k) k--;
    while (s[i + k] == s[sa[rank[i] - 1] + k]) k++;
    height[rank[i]] = k;
  }
  Log[0] = -1;
  for (int i = 1; i <= n; i++) Log[i] = Log[i >> 1] + 1;
  for (int i = 2; i <= n; i++) RMQ[i][0] = height[i];
  for (int j = 1; j <= 18; j++)
    for (int i = 2; i + (1 << j) - 1 <= n; i++)
      RMQ[i][j] = Min(RMQ[i][j - 1], RMQ[i + (1 << j - 1)][j - 1]);
}
int query(int l, int r) {
  int k = Log[r - l + 1];
  return Min(RMQ[l][k], RMQ[r - (1 << k) + 1][k]);
}
long long calc(int n, int m, int *a, int *b) {
  for (int i = 1; i <= n; i++) u[i] = rank[a[i]];
  for (int i = 1; i <= m; i++) v[i] = rank[b[i]] + 1;
  std::sort(u + 1, u + n + 1);
  std::sort(v + 1, v + m + 1);
  top = 0;
  long long res = 0, ans = 0;
  int p = 1;
  for (int i = 1; i <= n; i++) {
    int np = 0;
    while (top && query(u[i - 1] + 1, u[i]) <= val[top])
      res -= 1ll * cnt[top] * val[top], np += cnt[top--];
    if (np)
      cnt[++top] = np, res += 1ll * np * (val[top] = query(u[i - 1] + 1, u[i]));
    while (p <= m && v[p] <= u[i])
      cnt[++top] = 1, res += (val[top] = query(v[p], u[i])), p++;
    ans += res;
  }
  return ans;
}
int main() {
  n = read();
  q = read();
  scanf("%s", s + 1);
  init();
  while (q--) {
    k = read();
    l = read();
    for (int i = 1; i <= k; i++) a[i] = read();
    for (int i = 1; i <= l; i++) b[i] = read();
    for (int i = 1; i <= k; i++) vis[a[i]] = 1;
    long long res = 0;
    for (int i = 1; i <= l; i++)
      if (vis[b[i]]) res += n - b[i] + 1;
    for (int i = 1; i <= k; i++) vis[a[i]] = 0;
    printf("%lld\n", res + calc(k, l, a, b) + calc(l, k, b, a));
  }
  return 0;
}