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;
int n, m, id[200100], cnt = 1;
struct Suffix_Automaton {
  int ch[26], len, fa;
} t[400100];
int Add(int x, int c) {
  int xx = ++cnt;
  t[xx].len = t[x].len + 1;
  for (; x && !t[x].ch[c]; x = t[x].fa) t[x].ch[c] = xx;
  if (!x) {
    t[xx].fa = 1;
    return xx;
  }
  int y = t[x].ch[c];
  if (t[y].len == t[x].len + 1) {
    t[xx].fa = y;
    return xx;
  }
  int yy = ++cnt;
  t[yy] = t[y], t[yy].len = t[x].len + 1;
  t[xx].fa = t[y].fa = yy;
  for (; x && t[x].ch[c] == y; x = t[x].fa) t[x].ch[c] = yy;
  return xx;
}
char s[200100];
int fa[400100], dfn[400100], sz[400100], son[400100], rev[400100], top[400100],
    dep[400100], tot;
vector<int> v[400100];
void dfs1(int x) {
  sz[x] = 1;
  for (auto y : v[x]) {
    fa[y] = x, dep[y] = dep[x] + 1;
    dfs1(y);
    sz[x] += sz[y];
    if (sz[y] > sz[son[x]]) son[x] = y;
  }
}
void dfs2(int x) {
  dfn[x] = ++tot, rev[tot] = x;
  if (!top[x]) top[x] = x;
  if (son[x]) top[son[x]] = top[x], dfs2(son[x]);
  for (auto y : v[x])
    if (y != son[x]) dfs2(y);
}
struct BIT {
  long long s[400100];
  long long t1[400100], t2[400100];
  long long &operator[](const int &x) { return s[x]; }
  BIT() {
    memset(s, 0, sizeof(s)), memset(t1, 0, sizeof(t1)),
        memset(t2, 0, sizeof(t2));
  }
  void ADD(int x, int y) {
    for (int i = x; i <= cnt; i += i & -i)
      t1[i] += y, t2[i] += 1ll * s[x - 1] * y;
  }
  long long SUM(int x) {
    long long sum = 0;
    for (int i = x; i; i -= i & -i) sum += t1[i];
    sum *= s[x];
    for (int i = x; i; i -= i & -i) sum -= t2[i];
    return sum;
  }
  void ADD(int l, int r, int val) {
    if (l <= r) ADD(l, val), ADD(r + 1, -val);
  }
  long long SUM(int l, int r) { return SUM(r) - SUM(l - 1); }
} bit;
void chainadd(int x, int val) {
  x = id[n - x];
  while (x) bit.ADD(dfn[top[x]], dfn[x], val), x = fa[top[x]];
}
long long chainsum(int x) {
  x = id[n - x];
  long long ret = 0;
  while (x) ret += bit.SUM(dfn[top[x]], dfn[x]), x = fa[top[x]];
  return ret;
}
int a[200100];
long long res;
int read() {
  int x = 0;
  char c = getchar();
  while (c > '9' || c < '0') c = getchar();
  while (c >= '0' && c <= '9')
    x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  return x;
}
void print(long long x) {
  if (x <= 9)
    putchar('0' + x);
  else
    print(x / 10), putchar('0' + x % 10);
}
int main() {
  n = read(), m = read(), scanf("%s", s), reverse(s, s + n);
  for (int i = 0, las = 1; i < n; i++) id[i] = las = Add(las, s[i] - 'a');
  for (int i = 2; i <= cnt; i++) v[t[i].fa].push_back(i);
  dfs1(1), dfs2(1);
  for (int i = 1; i <= cnt; i++)
    bit[i] = bit[i - 1] + t[rev[i]].len - t[t[rev[i]].fa].len;
  for (int i = 1, A, B; i <= m; i++) {
    A = read(), B = read(), res = 0;
    for (int i = 1; i <= A; i++) chainadd(a[i] = read(), 1);
    for (int i = 1; i <= B; i++) res += chainsum(read());
    for (int i = 1; i <= A; i++) chainadd(a[i], -1);
    print(res), putchar('\n');
  }
  return 0;
}