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 inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int N = 2e5;
int n, m, an, bn, a[N + 7], b[N + 7];
long long ans;
char s[N + 7];
int sc, c[N + 7], tp[N + 7], rk[N + 7], sa[N + 7], h[N + 7];
void csort() {
  for (int i = 0; i <= sc; i++) c[i] = 0;
  for (int i = 1; i <= n; i++) c[rk[i]]++;
  for (int i = 1; i <= sc; i++) c[i] += c[i - 1];
  for (int i = n; i >= 1; i--) sa[c[rk[tp[i]]]--] = tp[i];
}
void buildsa() {
  for (int i = 1; i <= n; i++) rk[i] = s[i], tp[i] = i;
  sc = 128, csort();
  for (int w = 1, p = 0, i; p < n; w <<= 1, sc = p) {
    for (p = 0, i = n - w + 1; i <= n; i++) tp[++p] = i;
    for (i = 1; i <= n; i++)
      if (sa[i] > w) tp[++p] = sa[i] - w;
    csort(), swap(rk, tp), rk[sa[1]] = p = 1;
    for (i = 2; i <= n; rk[sa[i]] = p, i++)
      if (tp[sa[i]] != tp[sa[i - 1]] || tp[sa[i] + w] != tp[sa[i - 1] + w]) p++;
  }
  for (int i = 1, j, k = 0; i <= n; h[rk[i++]] = k)
    for (k = k ? k - 1 : k, j = sa[rk[i] - 1]; s[i + k] == s[j + k]; k++)
      ;
}
int st[20][N + 7];
void buildst() {
  for (int i = 1; i <= n; i++) st[0][i] = h[i];
  for (int j = 1; j <= 18; j++)
    for (int i = 1; i + (1 << j) - 1 <= n; i++)
      st[j][i] = min(st[j - 1][i], st[j - 1][i + (1 << (j - 1))]);
}
int stget(int l, int r) {
  int lg = log2(r - l + 1);
  return min(st[lg][l], st[lg][r - (1 << lg) + 1]);
}
int lcp(int l, int r) {
  if (l == r) return n - l + 1;
  l = rk[l], r = rk[r];
  if (l > r) l ^= r ^= l ^= r;
  return stget(l + 1, r);
}
const int T = N << 2;
struct node {
  int l, r, mx, mn;
  long long sm;
  node(int lp = 0, int rp = 0, int xm = 0, int nm = 0, long long ms = 0) {
    l = lp, r = rp, mx = xm, mn = nm, sm = ms;
  }
  node operator+(const node q) {
    return node(l, q.r, max(mx, q.mx), min(mn, q.mn), sm + q.sm);
  }
};
int mk[T + 7];
node v[T + 7];
void pushup(int k) { v[k] = v[k << 1] + v[k << 1 | 1]; }
void uni(int k, int first) {
  v[k].mn = v[k].mx = first, v[k].sm = (long long)first * (v[k].r - v[k].l + 1),
  mk[k] = first;
}
void pushdown(int k) {
  if (~mk[k]) uni(k << 1, mk[k]), uni(k << 1 | 1, mk[k]), mk[k] = -1;
}
void buildsm(int k, int first, int second) {
  mk[k] = -1, v[k].l = first, v[k].r = second;
  if (first == second) return uni(k, 0);
  buildsm(k << 1, first, ((first + second) >> 1)),
      buildsm(k << 1 | 1, ((first + second) >> 1) + 1, second), pushup(k);
}
void press(int k, int first) {
  if (v[k].mx <= first) return;
  if (v[k].mn >= first) return mk[k] = first, uni(k, first);
  pushdown(k), press(k << 1, first), press(k << 1 | 1, first), pushup(k);
}
void fix(int k, int first, int second) {
  if (v[k].l == v[k].r) return uni(k, second);
  pushdown(k);
  if (((v[k].l + v[k].r) >> 1) >= first)
    fix(k << 1, first, second);
  else
    fix(k << 1 | 1, first, second);
  pushup(k);
}
int main() {
  scanf("%d%d%s", &n, &m, &s[1]);
  buildsa(), buildst();
  for (int mi = 1; mi <= m; mi++) {
    scanf("%d%d", &an, &bn), ans = 0;
    for (int i = 1; i <= an; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= bn; i++) scanf("%d", &b[i]);
    sort(a + 1, a + an + 1,
         [&](int first, int second) { return rk[first] < rk[second]; });
    sort(b + 1, b + bn + 1,
         [&](int first, int second) { return rk[first] < rk[second]; });
    buildsm(1, 1, an);
    for (int i = 1, j = 1; j <= bn; j++) {
      press(1, lcp(b[j - 1], b[j]));
      for (; i <= an && rk[a[i]] <= rk[b[j]]; i++) fix(1, i, lcp(a[i], b[j]));
      ans += v[1].sm;
    }
    buildsm(1, 1, an);
    for (int i = an, j = bn; j >= 1; j--) {
      press(1, lcp(b[j + 1], b[j]));
      for (; i >= 1 && rk[a[i]] > rk[b[j]]; i--) fix(1, i, lcp(a[i], b[j]));
      ans += v[1].sm;
    }
    printf("%lld\n", ans);
  }
  return 0;
}