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;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar(), f = 0;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = 1;
  for (; c >= '0' && c <= '9'; c = getchar())
    x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
}
const int N = 200005, Z = 26;
vector<int> v[N << 1];
int ru[N << 1], rut;
struct node {
  int ch[Z], len, fa;
} S[N << 1];
int sls, smt, fa[N << 1][19], dep[N << 1];
int id[N], dt, dfn[N << 1], s1[N << 1], s2[N << 1], st[N], tp, n, Q;
long long rs = 0;
struct edge {
  int to, nxt;
} e[N << 2];
int et, head[N << 1], at, a[N << 1];
char ch[N];
inline void adde(int x, int y) { e[++et] = (edge){y, head[x]}, head[x] = et; }
inline char cmp(int a, int b) { return dfn[a] < dfn[b]; }
inline void saminit() { sls = smt = 1; }
inline void samins(int c) {
  int p = sls, np = sls = ++smt;
  S[np].len = S[p].len + 1;
  for (; p && !S[p].ch[c]; p = S[p].fa) S[p].ch[c] = np;
  int q = S[p].ch[c];
  if (!p) return S[np].fa = 1, void();
  if (S[p].len + 1 == S[q].len) return S[np].fa = q, void();
  int nq = ++smt;
  S[nq] = S[q], S[nq].len = S[p].len + 1, S[q].fa = S[np].fa = nq;
  for (; p && S[p].ch[c] == q; p = S[p].fa) S[p].ch[c] = nq;
}
inline void dfs0(int x, int f) {
  dep[x] = dep[f] + 1, dfn[x] = ++dt, fa[x][0] = f;
  for (int i = 1; i < 19; i++) fa[x][i] = fa[fa[x][i - 1]][i - 1];
  for (auto i : v[x]) dfs0(i, x);
}
inline void init() {
  for (int i = 2; i <= smt; i++) v[S[i].fa].push_back(i);
  dfs0(1, 0);
  for (int i = n, p = 1; i; i--) id[i] = p = S[p].ch[ch[i] - 'a'];
}
inline int LCA(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (int d = dep[x] - dep[y], i = 0; d; d >>= 1, i++)
    if (d & 1) x = fa[x][i];
  if (x == y) return x;
  for (int i = 18; ~i; i--)
    if (fa[x][i] ^ fa[y][i]) x = fa[x][i], y = fa[y][i];
  return fa[x][0];
}
inline void dfs(int x, int fa) {
  int tmp = s1[x];
  for (int i = head[x]; i; i = e[i].nxt)
    if (e[i].to != fa) dfs(e[i].to, x);
  for (int i = head[x]; i; i = e[i].nxt)
    if (e[i].to != fa) s1[x] += s1[e[i].to], s2[x] += s2[e[i].to];
  for (int i = head[x]; i; i = e[i].nxt)
    if (e[i].to != fa)
      rs += 1ll * s1[e[i].to] * (s2[x] - s2[e[i].to]) * S[x].len;
  rs += 1ll * tmp * s2[x] * S[x].len;
}
int main() {
  saminit(), read(n), read(Q), scanf("%s", ch + 1);
  for (int i = n; i; i--) samins(ch[i] - 'a');
  init();
  for (int p, q; Q--;) {
    read(p), read(q), at = 0, st[tp = 1] = 1, head[1] = 0, et = 0, rut = 0;
    for (int i = 1, x; i <= p; i++) read(x), x = id[x], s1[x]++, a[++at] = x;
    for (int i = 1, x; i <= q; i++) read(x), x = id[x], s2[x]++, a[++at] = x;
    sort(a + 1, a + at + 1, cmp), at = unique(a + 1, a + at + 1) - a - 1;
    for (int i = 1; i <= at; i++) ru[++rut] = a[i];
    for (int i = 1; i <= at; a[i] ^ 1 ? st[++tp] = a[i++] : ++i) {
      if (a[i] == 1)
        continue;
      else
        head[a[i]] = 0;
      int lc = LCA(a[i], st[tp]);
      if (lc == st[tp])
        continue;
      else
        ru[++rut] = lc;
      while (dfn[lc] < dfn[st[tp - 1]])
        adde(st[tp - 1], st[tp]), adde(st[tp], st[tp - 1]), tp--;
      if (dfn[lc] == dfn[st[tp - 1]])
        adde(lc, st[tp]), adde(st[tp], lc), tp--;
      else
        head[lc] = 0, adde(lc, st[tp]), adde(st[tp], lc), st[tp] = lc;
    }
    for (int i = 1; i < tp; i++) adde(st[i], st[i + 1]), adde(st[i + 1], st[i]);
    rs = 0, dfs(1, 0), printf("%lld\n", rs);
    for (int i = 1; i <= rut; i++) s1[ru[i]] = s2[ru[i]] = 0;
  }
  return 0;
}