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;
char s[200005];
int n, T, m, sa[400005], rk[400005], tp[400005], tax[400005], h[400005],
    height[400005];
struct SGT {
  int tree[1000005];
  SGT() {
    memset(tree, 0x7f, sizeof(tree));
    return;
  }
  inline void pushup(int now) {
    tree[now] = min(tree[now * 2], tree[now * 2 + 1]);
    return;
  }
  inline void tree_update(int now, int nowl, int nowr, int pos, int val) {
    if (nowl == nowr) {
      tree[now] = val;
      return;
    }
    int mid = (nowl + nowr) / 2;
    if (pos <= mid)
      tree_update(now * 2, nowl, mid, pos, val);
    else
      tree_update(now * 2 + 1, mid + 1, nowr, pos, val);
    pushup(now);
    return;
  }
  inline void update(int pos, int val) {
    tree_update(1, 1, n, pos, val);
    return;
  }
  inline int tree_query(int now, int nowl, int nowr, int lt, int rt) {
    if (nowl >= lt && nowr <= rt) return tree[now];
    int mid = (nowl + nowr) / 2, s = 1e9;
    if (mid >= lt) s = min(s, tree_query(now * 2, nowl, mid, lt, rt));
    if (mid + 1 <= rt)
      s = min(s, tree_query(now * 2 + 1, mid + 1, nowr, lt, rt));
    return s;
  }
  inline int query(int l, int r) { return tree_query(1, 1, n, l, r); }
} sgt;
inline void Qsort() {
  memset(tax, 0, sizeof(tax));
  for (int i = 1; i <= n; i++) tax[rk[i]]++;
  for (int i = 1; i <= m; i++) tax[i] += tax[i - 1];
  for (int i = n; i >= 1; i--) sa[tax[rk[tp[i]]]--] = tp[i];
  return;
}
inline void Suffix_Sorting() {
  m = 200;
  for (int i = 1; i <= n; i++) tp[i] = i, rk[i] = (int)(s[i]);
  Qsort();
  for (int w = 1, p = 0; p < n; m = p, w *= 2) {
    p = 0;
    for (int i = n - w + 1; i <= n; i++) tp[++p] = i;
    for (int i = 1; i <= n; i++)
      if (sa[i] > w) tp[++p] = sa[i] - w;
    Qsort();
    swap(rk, tp);
    rk[sa[1]] = 1;
    p = 1;
    for (int i = 2; i <= n; i++) {
      if (tp[sa[i]] == tp[sa[i - 1]] && tp[sa[i] + w] == tp[sa[i - 1] + w])
        rk[sa[i]] = p;
      else
        rk[sa[i]] = ++p;
    }
  }
  return;
}
inline void Get_Height() {
  for (int i = 1; i <= n; i++) {
    h[i] = max(0, h[i - 1] - 1);
    while (s[i + h[i]] == s[sa[rk[i] - 1] + h[i]] && i + h[i] <= n &&
           sa[rk[i] - 1] + h[i] <= n)
      h[i]++;
  }
  for (int i = 1; i <= n; i++) height[i] = h[sa[i]];
  return;
}
int a[200005], b[200005], c[400005], tot;
int tail;
struct node {
  int cnt, val;
  node() { cnt = val = 0; }
} q[400005];
long long calc() {
  for (int i = 1; i <= tot; i++) c[i] = rk[c[i]];
  sort(c + 1, c + 1 + tot);
  if (tot == 1) return 0;
  tot--;
  for (int i = 1; i <= tot; i++)
    if (c[i] == c[i + 1])
      c[i] = n - sa[c[i]] + 1;
    else
      c[i] = sgt.query(c[i] + 1, c[i + 1]);
  long long nowsum = 0, sum = 0;
  tail = 0;
  for (int i = 1; i <= tot; i++) {
    node now;
    now.cnt = 1, now.val = c[i];
    while (tail > 0 && q[tail].val > now.val) {
      now.cnt += q[tail].cnt;
      nowsum -= q[tail].cnt * q[tail].val;
      tail--;
    }
    q[++tail] = now;
    nowsum += q[tail].cnt * q[tail].val;
    sum += nowsum;
  }
  return sum;
}
int main() {
  cin >> n >> T;
  scanf("%s", s + 1);
  Suffix_Sorting();
  Get_Height();
  for (int i = 2; i <= n; i++) sgt.update(i, height[i]);
  while (T--) {
    int l1, l2;
    scanf("%d%d", &l1, &l2);
    for (int i = 1; i <= l1; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= l2; i++) scanf("%d", &b[i]);
    long long ans = 0;
    tot = 0;
    for (int i = 1; i <= l1; i++) c[++tot] = a[i];
    ans -= calc();
    tot = 0;
    for (int i = 1; i <= l2; i++) c[++tot] = b[i];
    ans -= calc();
    tot = 0;
    for (int i = 1; i <= l1; i++) c[++tot] = a[i];
    for (int i = 1; i <= l2; i++) c[++tot] = b[i];
    ans += calc();
    printf("%lld\n", ans);
  }
}