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();
  bool flag = false;
  while (!isdigit(c)) {
    if (c == '-') flag = true;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  if (flag) x = -x;
}
int n, q, na, nb, las, root, tot, top, dfn_cnt, num, last;
long long ans;
int fa[400010], len[400010], ch[400010][26], bel[200010], qu[400010];
int dfn[400010], siz[400010], son[400010], de[400010], tp[400010], st[400010];
long long tag[400010][2], f[400010][2];
char s[200010];
struct edge {
  int to, nxt;
} e[400010];
int head[400010], edge_cnt;
void add(int from, int to) {
  e[++edge_cnt] = (edge){to, head[from]};
  head[from] = edge_cnt;
}
bool cmp(const int &a, const int &b) { return dfn[a] < dfn[b]; }
void init() { las = root = tot = 1; }
void insert_sam(int c, int pos) {
  int p = las, np = las = ++tot;
  len[np] = len[p] + 1, bel[pos] = np;
  while (p && !ch[p][c]) ch[p][c] = np, p = fa[p];
  if (!p)
    fa[np] = root;
  else {
    int q = ch[p][c];
    if (len[q] == len[p] + 1)
      fa[np] = q;
    else {
      int nq = ++tot;
      memcpy(ch[nq], ch[q], sizeof(ch[q]));
      len[nq] = len[p] + 1, fa[nq] = fa[q], fa[q] = fa[np] = nq;
      while (ch[p][c] == q) ch[p][c] = nq, p = fa[p];
    }
  }
}
void dfs_son(int x) {
  dfn[x] = ++dfn_cnt, siz[x] = 1, de[x] = de[fa[x]] + 1;
  for (int i = head[x]; i; i = e[i].nxt) {
    int y = e[i].to;
    if (y == fa[x]) continue;
    dfs_son(y), siz[x] += siz[y];
    if (siz[son[x]] < siz[y]) son[x] = y;
  }
}
void dfs_chain(int x, int t) {
  tp[x] = t;
  if (son[x]) dfs_chain(son[x], t);
  for (int i = head[x]; i; i = e[i].nxt) {
    int y = e[i].to;
    if (tp[y]) continue;
    dfs_chain(y, y);
  }
}
int lca(int x, int y) {
  while (tp[x] != tp[y]) {
    if (de[tp[x]] < de[tp[y]]) swap(x, y);
    x = fa[tp[x]];
  }
  if (de[x] > de[y]) swap(x, y);
  return x;
}
void insert_tree(int x) {
  if (x == root) return;
  if (top == 1) {
    st[++top] = x;
    return;
  }
  int anc = lca(x, st[top]);
  if (anc == st[top]) {
    st[++top] = x;
    return;
  }
  while (top > 1 && dfn[anc] <= dfn[st[top - 1]])
    add(st[top - 1], st[top]), top--;
  if (anc != st[top]) add(anc, st[top]), st[top] = anc;
  st[++top] = x;
}
void dfs_dp(int x) {
  for (int i = head[x]; i; i = e[i].nxt) {
    int y = e[i].to;
    dfs_dp(y), f[x][0] += f[y][0], f[x][1] += f[y][1];
  }
  for (int i = head[x]; i; i = e[i].nxt) {
    int y = e[i].to;
    ans += (f[x][0] - f[y][0]) * f[y][1] * len[x];
  }
  ans += (tag[x][0] * tag[x][1] + tag[x][0] * f[x][1] + tag[x][1] * f[x][0]) *
         len[x];
  f[x][0] += tag[x][0], f[x][1] += tag[x][1];
}
void clear() { num = last = ans = edge_cnt = 0, st[top = 1] = 1; }
void del(int x) {
  tag[x][0] = tag[x][1] = f[x][0] = f[x][1] = 0;
  for (int i = head[x]; i; i = e[i].nxt) del(e[i].to);
  head[x] = 0;
}
int main() {
  init();
  read(n), read(q), scanf("%s", s + 1), reverse(s + 1, s + n + 1);
  for (int i = 1; i <= n; ++i) insert_sam(s[i] - 'a', n - i + 1);
  for (int i = 2; i <= tot; ++i) add(fa[i], i);
  dfs_son(root), dfs_chain(root, root);
  edge_cnt = 0, memset(head, 0, sizeof(head));
  while (q--) {
    clear();
    read(na), read(nb);
    for (int i = 1; i <= na; ++i) {
      int p;
      read(p), p = bel[p];
      tag[p][0] = 1, qu[++num] = p;
    }
    for (int i = 1; i <= nb; ++i) {
      int p;
      read(p), p = bel[p];
      tag[p][1] = 1, qu[++num] = p;
    }
    sort(qu + 1, qu + num + 1, cmp);
    for (int i = 1; i <= num; ++i) {
      if (qu[i] == last) continue;
      insert_tree(qu[i]), last = qu[i];
    }
    while (top) add(st[top - 1], st[top]), top--;
    dfs_dp(root), del(root);
    printf("%lld\n", ans);
  }
  return 0;
}