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;
long long n, m, vis[605010], st[605010], rnk[605010], sa[605010], tp[605010];
long long read() {
  long long x;
  scanf("%lld", &x);
  return x;
}
long long height[605010], s[605010];
char dou[605010];
inline void qsort() {
  for (long long i = 1; i <= m + 233; ++i) st[i] = 0;
  for (long long i = 1; i <= n; ++i) st[rnk[i]]++;
  for (long long i = 1; i <= m + 233; ++i) st[i] = st[i] + st[i - 1];
  for (long long i = n; i >= 1; --i) sa[st[rnk[tp[i]]]--] = tp[i];
}
inline void get_height() {
  long long j, k = 0;
  for (long long i = 1; i <= n; ++i) {
    if (k) k--;
    j = sa[rnk[i] - 1];
    while (dou[i + k] == dou[j + k] && i + k <= n && j + k <= n) k++;
    height[rnk[i]] = k;
  }
}
long long dp[605010][23];
long long lcp(long long x, long long y) {
  long long L = rnk[x], R = rnk[y];
  if (L > R) swap(L, R);
  if (L == R) return n - x + 1;
  L = L + 1;
  long long tmp = log2(R - L + 1);
  return min(dp[L][tmp], dp[R - (1 << tmp) + 1][tmp]);
}
long long a[605010], b[605010], sizea[605010], sizeb[605010], cao[605010];
struct node {
  long long rnk, who, opt;
  bool operator<(const node& a) const { return rnk < a.rnk; }
} init[605010];
long long sta[605010], top;
long long len[605010], tot;
inline void work() {
  long long res = 0;
  long long numa = read(), numb = read(), num = 0;
  for (long long i = 1; i <= numa; ++i)
    a[i] = read(), init[++num] = (node){rnk[a[i]], a[i], 1};
  for (long long i = 1; i <= numb; ++i)
    b[i] = read(), init[++num] = (node){rnk[b[i]], b[i], 2};
  sort(init + 1, init + 1 + num);
  tot = n + 1;
  top = 0;
  sta[++top] = init[1].who;
  cao[sta[top]] = 0;
  if (init[1].opt == 1)
    sizea[sta[top]] = 1, sizeb[sta[top]] = 0;
  else
    sizeb[sta[top]] = 1, sizea[sta[top]] = 0;
  len[sta[top]] = n - init[1].who + 1;
  for (long long i = 2; i <= num; ++i) {
    long long who = init[i].who, opt = init[i].opt;
    if (init[i].who == sta[top]) {
      if (opt == 1)
        sizea[who]++;
      else
        sizeb[who]++;
      cao[sta[top]]++;
      continue;
    }
    while (top >= 2 && lcp(init[i - 1].who, who) <= len[sta[top - 1]]) {
      res = res + cao[sta[top]] * len[sta[top]];
      long long x = sta[top], xx = sta[top - 1];
      cao[xx] += sizea[xx] * sizeb[x] + sizeb[xx] * sizea[x];
      sizea[xx] += sizea[x];
      sizeb[xx] += sizeb[x];
      top--;
    }
    if (len[sta[top]] > lcp(init[i - 1].who, who)) {
      tot++;
      len[tot] = lcp(init[i - 1].who, who);
      sizea[tot] = sizeb[tot] = cao[tot] = 0;
      res = res + cao[sta[top]] * len[sta[top]];
      long long x = sta[top], xx = tot;
      cao[xx] += sizea[xx] * sizeb[x] + sizeb[xx] * sizea[x];
      sizea[xx] += sizea[x];
      sizeb[xx] += sizeb[x];
      top--;
      sta[++top] = tot;
    }
    len[who] = n - who + 1;
    cao[who] = 0;
    if (opt == 1)
      sizea[who] = 1, sizeb[who] = 0;
    else
      sizeb[who] = 1, sizea[who] = 0;
    sta[++top] = who;
  }
  while (top >= 2) {
    res = res + cao[sta[top]] * len[sta[top]];
    long long x = sta[top], xx = sta[top - 1];
    cao[xx] += sizea[xx] * sizeb[x] + sizeb[xx] * sizea[x];
    sizea[xx] += sizea[x];
    sizeb[xx] += sizeb[x];
    top--;
  }
  if (top == 1) res = res + cao[sta[top]] * len[sta[top]];
  printf("%lld\n", res);
}
long long M;
signed main() {
  n = read(), M = read();
  cin >> (dou + 1);
  m = n;
  for (long long i = 1; i <= n; ++i) rnk[i] = dou[i], tp[i] = i;
  qsort();
  long long top;
  for (long long len = 1;; len <<= 1) {
    top = 0;
    for (long long i = 1; i <= len; ++i) tp[++top] = n - i + 1;
    for (long long i = 1; i <= n; ++i)
      if (sa[i] > len) tp[++top] = sa[i] - len;
    qsort();
    for (long long i = 1; i <= n; ++i) tp[i] = rnk[i];
    top = 0;
    for (long long i = 1; i <= n; ++i)
      rnk[sa[i]] =
          (tp[sa[i]] == tp[sa[i - 1]] && tp[sa[i] + len] == tp[sa[i - 1] + len])
              ? top
              : ++top;
    if (top == n) break;
    m = top;
  }
  get_height();
  for (long long i = 1; i <= n; ++i) dp[i][0] = height[i];
  for (long long k = 1; k <= 22; ++k)
    for (long long i = 1; i <= n; ++i)
      if (i + (1 << (k - 1)) <= n)
        dp[i][k] = min(dp[i][k - 1], dp[i + (1 << (k - 1))][k - 1]);
  while (M--) work();
  return 0;
}