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 N = 5e5 + 10;
int n, m, tot, Siz, type, cnt1, cnt2, cnt, top, last, x;
int head[N], dep[N], Top[N], fa[N], siz[N], son[N], st[N], b[N], num[N][2],
    dfn[N], sta[N];
int link[N], len[N], tr[N][30];
long long ans;
char s[N];
vector<int> v[N];
struct node {
  int to, net;
} e[N << 1];
inline int read() {
  int s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * w;
}
void add(int x, int y) {
  e[++tot].to = y;
  e[tot].net = head[x];
  head[x] = tot;
}
void Add(int x, int y) { v[x].push_back(y); }
bool comp(int a, int b) { return dfn[a] < dfn[b]; }
void Extend(int ch) {
  int now = ++Siz, p;
  len[now] = len[last] + 1;
  for (p = last; p && !tr[p][ch]; p = link[p]) tr[p][ch] = now;
  if (!p)
    link[now] = 1;
  else {
    int x = tr[p][ch];
    if (len[x] == len[p] + 1)
      link[now] = x;
    else {
      int y = ++Siz;
      len[y] = len[p] + 1;
      memcpy(tr[y], tr[x], sizeof(tr[x]));
      link[y] = link[x];
      link[x] = link[now] = y;
      while (p && tr[p][ch] == x) {
        tr[p][ch] = y;
        p = link[p];
      }
    }
  }
  last = now;
}
void get_tree(int x) {
  dep[x] = dep[fa[x]] + 1;
  siz[x] = 1;
  for (int i = head[x]; i; i = e[i].net) {
    int to = e[i].to;
    if (to == fa[x]) continue;
    fa[to] = x;
    get_tree(to);
    siz[x] += siz[to];
    if (siz[to] > siz[son[x]]) son[x] = to;
  }
}
void dfs(int x, int topp) {
  Top[x] = topp;
  dfn[x] = ++type;
  if (son[x]) dfs(son[x], topp);
  for (int i = head[x]; i; i = e[i].net) {
    int to = e[i].to;
    if (to == fa[x] || to == son[x]) continue;
    dfs(to, to);
  }
}
int Lca(int x, int y) {
  while (Top[x] != Top[y]) {
    if (dep[Top[x]] < dep[Top[y]]) swap(x, y);
    x = fa[Top[x]];
  }
  return dep[x] <= dep[y] ? x : y;
}
void build() {
  sta[++top] = 1;
  int num = unique(b + 1, b + cnt + 1) - b - 1;
  for (int i = 1; i <= num; i++) {
    int x = b[i];
    int lca = Lca(x, sta[top]);
    while (top > 1 && dep[lca] <= dep[sta[top - 1]])
      Add(sta[top - 1], sta[top]), top--;
    if (sta[top] != lca) Add(lca, sta[top]), sta[top] = lca;
    if (sta[top] != x) sta[++top] = x;
  }
  while (top > 1) Add(sta[top - 1], sta[top]), top--;
}
void dp(int x, int fa) {
  ans += 1LL * len[x] * num[x][0] * num[x][1];
  for (int i = 0; i < v[x].size(); i++) {
    int to = v[x][i];
    if (to == fa) continue;
    dp(to, x);
    ans += 1LL * len[x] *
           (1LL * num[x][0] * num[to][1] + 1LL * num[x][1] * num[to][0]);
    num[x][0] += num[to][0];
    num[x][1] += num[to][1];
  }
}
void QK(int x, int fa) {
  num[x][0] = num[x][1] = 0;
  for (int i = 0; i < v[x].size(); i++) {
    int to = v[x][i];
    if (to == fa) continue;
    QK(to, x);
  }
  v[x].clear();
}
int main() {
  n = read();
  m = read();
  scanf("%s", s + 1);
  last = Siz = 1;
  for (int i = n; i >= 1; i--) Extend(s[i] - 'a'), st[i] = last;
  for (int i = 2; i <= Siz; i++) add(link[i], i);
  get_tree(1);
  dfs(1, 0);
  for (int i = 1; i <= m; i++) {
    cnt1 = read();
    cnt2 = read();
    cnt = top = 0;
    for (int j = 1; j <= cnt1; j++) {
      x = read();
      b[++cnt] = st[x];
      num[st[x]][0]++;
    }
    for (int j = 1; j <= cnt2; j++) {
      x = read();
      b[++cnt] = st[x];
      num[st[x]][1]++;
    }
    sort(b + 1, b + cnt + 1, comp);
    build();
    dp(1, 1);
    QK(1, 1);
    printf("%lld\n", ans);
    ans = 0;
  }
  return 0;
}