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 MAXN = 4e5 + 10;
int n, q, clo, now, cnt, qcnt, nodecnt;
char str[MAXN];
int id[MAXN], father[MAXN][20], lst[MAXN], dep[MAXN], flag[MAXN], dfn[MAXN],
    sta[MAXN], len[MAXN], fa[MAXN], fb[MAXN], que[MAXN];
vector<int> e[MAXN], G[MAXN];
struct node {
  int nx[26], len, root;
} T[MAXN];
void init() {
  nodecnt = 0;
  T[0].root = -1, T[0].len = 0;
}
int extend(int p, int c) {
  int np = ++nodecnt;
  T[np].len = T[p].len + 1;
  for (; p != -1 && !T[p].nx[c]; p = T[p].root) T[p].nx[c] = np;
  if (p == -1)
    T[np].root = 0;
  else {
    int q = T[p].nx[c];
    if (T[q].len == T[p].len + 1)
      T[np].root = q;
    else {
      int nq = ++nodecnt;
      T[nq] = T[q];
      T[nq].len = T[p].len + 1;
      for (; p != -1 && T[p].nx[c] == q; p = T[p].root) T[p].nx[c] = nq;
      T[q].root = T[np].root = nq;
    }
  }
  return np;
}
void dfs(int x) {
  dfn[x] = ++clo;
  for (auto v : e[x]) {
    father[v][0] = x;
    dep[v] = dep[x] + 1;
    dfs(v);
  }
}
int lca(int x, int y) {
  int i, j;
  if (dep[x] < dep[y]) swap(x, y);
  for (i = 0; (1 << i) <= dep[x]; ++i)
    ;
  --i;
  for (j = i; j >= 0; --j) {
    if (dep[x] - (1 << j) >= dep[y]) {
      x = father[x][j];
    }
  }
  if (x == y) return x;
  for (j = i; j >= 0; --j) {
    if (father[x][j] != father[y][j]) {
      x = father[x][j], y = father[y][j];
    }
  }
  return father[x][0];
}
bool cmp(const int &a, const int &b) { return dfn[a] < dfn[b]; }
int ed = 0;
void addedge(int x, int y) {
  ++ed;
  G[x].push_back(y);
  if (flag[x] != now) {
    flag[x] = now;
    que[++qcnt] = x;
  }
  if (flag[y] != now) {
    flag[y] = now;
    que[++qcnt] = y;
  }
}
long long ans = 0;
long long dp(int x) {
  long long tmp = 0;
  for (auto v : G[x]) {
    tmp += dp(v);
    fa[x] += fa[v];
    fb[x] += fb[v];
  }
  ans += 1LL * T[x].len * (1LL * fa[x] * fb[x] - tmp);
  return 1LL * fa[x] * fb[x];
}
void build(int a, int b) {
  ++now;
  int stm = 0;
  sort(lst + 1, lst + cnt + 1, cmp);
  cnt = unique(lst + 1, lst + cnt + 1) - lst - 1;
  sta[stm = 1] = lst[1];
  for (int i = 2, x; i <= cnt; ++i) {
    x = lst[i];
    int lc = lca(x, sta[stm]);
    for (; stm > 1 && dep[sta[stm - 1]] > dep[lc]; stm--) {
      addedge(sta[stm - 1], sta[stm]);
    }
    if (stm && dep[sta[stm]] > dep[lc]) {
      addedge(lc, sta[stm--]);
    }
    if (!stm || sta[stm] != lc) sta[++stm] = lc;
    sta[++stm] = x;
  }
  for (; stm > 1; --stm) addedge(sta[stm - 1], sta[stm]);
  dp(sta[1]);
  printf("%lld\n", ans);
}
int main() {
  scanf("%d%d", &n, &q);
  scanf("%s", str + 1);
  init();
  for (int i = n, p = 0; i > 0; --i) {
    p = extend(p, str[i] - 'a');
    id[i] = p;
    len[p] = n - i + 1;
  }
  for (int i = 1; i <= nodecnt; ++i) {
    e[T[i].root].push_back(i);
  }
  dfs(0);
  for (int j = 1; (1 << j) <= nodecnt + 1; ++j) {
    for (int i = 0; i <= nodecnt; ++i) {
      father[i][j] = father[father[i][j - 1]][j - 1];
    }
  }
  for (int i = 1; i <= q; ++i) {
    int x, a, b;
    scanf("%d%d", &a, &b);
    cnt = qcnt = 0;
    ans = 0;
    int aa, bb;
    for (int i = 1; i <= a; ++i) {
      scanf("%d", &x);
      aa = x;
      fa[id[x]]++;
      lst[++cnt] = id[x];
    }
    for (int i = 1; i <= b; ++i) {
      scanf("%d", &x);
      bb = x;
      fb[id[x]]++;
      lst[++cnt] = id[x];
    }
    build(aa, bb);
    for (int i = 1; i <= qcnt; ++i) {
      G[que[i]].clear();
      fa[que[i]] = fb[que[i]] = 0;
    }
    for (int i = 1; i <= cnt; ++i) {
      G[lst[i]].clear();
      fa[lst[i]] = fb[lst[i]] = 0;
    }
  }
  return 0;
}