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 = 2 * 2e5 + 100;
char s[maxn];
int f[maxn][21], dfn[maxn], dfn_cnt, dep[maxn];
vector<int> son[maxn], side[maxn];
struct SAM {
  int len[maxn], ch[maxn][26], link[maxn], last, cnt;
  SAM() { last = cnt = 1; }
  int insert(int c) {
    int cur = ++cnt, p = last;
    len[cur] = len[last] + 1;
    while (p && !ch[p][c]) ch[p][c] = cur, p = link[p];
    if (!p)
      link[cur] = 1;
    else {
      int q = ch[p][c];
      if (len[p] + 1 == len[q])
        link[cur] = q;
      else {
        int cl = ++cnt;
        len[cl] = len[p] + 1;
        memcpy(ch[cl], ch[q], sizeof(ch[cl]));
        link[cl] = link[q];
        link[cur] = link[q] = cl;
        while (p && ch[p][c] == q) ch[p][c] = cl, p = link[p];
      }
    }
    last = cur;
    return cur;
  }
  void add() {
    for (int i = (2), iend = (cnt); i <= iend; i++) {
      son[link[i]].push_back(i);
    }
  }
} t;
int n, q;
int loc[maxn];
int LCA(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  for (int i = 20; i >= 0; i--)
    if (dep[f[u][i]] >= dep[v]) u = f[u][i];
  if (u == v) return u;
  for (int i = 20; i >= 0; i--)
    if (f[u][i] != f[v][i]) u = f[u][i], v = f[v][i];
  return f[u][0];
}
void dfs1(int u, int fa) {
  dfn[u] = ++dfn_cnt;
  dep[u] = dep[fa] + 1;
  f[u][0] = fa;
  for (int i = (1), iend = (20); i <= iend; i++)
    f[u][i] = f[f[u][i - 1]][i - 1];
  for (int i = 0; i < son[u].size(); i++) {
    int v = son[u][i];
    if (v == fa) continue;
    dfs1(v, u);
  }
}
int sz1[maxn], sz2[maxn];
int tag1[maxn], tag2[maxn];
long long ans;
void dfs2(int u, int fa) {
  int buc1 = tag1[u], buc2 = tag2[u];
  long long tmp = 0;
  tmp += 1ll * buc1 * buc2 * t.len[u];
  ans += 1ll * buc1 * buc2 * t.len[u];
  for (int i = 0; i < side[u].size(); i++) {
    int v = side[u][i];
    if (v == fa) continue;
    dfs2(v, u);
    buc1 += sz1[v];
    buc2 += sz2[v];
  }
  if (u == 2) {
    int ggg = 1;
  }
  ans += 1ll * tag1[u] * (buc2 - tag2[u]) * t.len[u];
  ans += 1ll * tag2[u] * (buc1 - tag1[u]) * t.len[u];
  for (int i = 0; i < side[u].size(); i++) {
    int v = side[u][i];
    if (v == fa) continue;
    tmp += 1ll * sz1[v] * (buc2 - sz2[v] - tag2[u]) * t.len[u];
    ans += 1ll * sz1[v] * (buc2 - sz2[v] - tag2[u]) * t.len[u];
  }
  sz1[u] = buc1;
  sz2[u] = buc2;
}
vector<int> s1, s2;
bool inbuc[maxn];
vector<int> mem;
bool cmp(int a, int b) { return dfn[a] < dfn[b]; }
int stk[maxn];
int build() {
  int top = 1;
  stk[top] = mem[0];
  for (int i = 1; i < mem.size(); i++) {
    int u = mem[i];
    while (LCA(u, stk[top]) != stk[top]) top--;
    side[stk[top]].push_back(u);
    stk[++top] = u;
  }
  return mem[0];
}
void clear() {
  for (int i = 0; i < mem.size(); i++) {
    int u = mem[i];
    tag1[u] = tag2[u] = inbuc[u] = 0;
    side[u].clear();
  }
}
int main() {
  scanf("%d%d", &n, &q);
  scanf("%s", s + 1);
  reverse(s + 1, s + 1 + n);
  for (int i = (1), iend = (n); i <= iend; i++) {
    loc[i] = t.insert(s[i] - 'a');
  }
  t.add();
  dfs1(1, 0);
  for (int i = (1), iend = (q); i <= iend; i++) {
    s1.clear();
    s2.clear();
    mem.clear();
    int t1, t2;
    scanf("%d%d", &t1, &t2);
    for (int j = (1), jend = (t1); j <= jend; j++) {
      int idx;
      scanf("%d", &idx);
      idx = n - idx + 1;
      idx = loc[idx];
      if (!inbuc[idx]) inbuc[idx] = 1, mem.push_back(idx);
      tag1[idx] = 1;
      s1.push_back(idx);
    }
    for (int j = (1), jend = (t2); j <= jend; j++) {
      int idx;
      scanf("%d", &idx);
      idx = n - idx + 1;
      idx = loc[idx];
      if (!inbuc[idx]) inbuc[idx] = 1, mem.push_back(idx);
      tag2[idx] = 1;
      s2.push_back(idx);
    }
    sort(mem.begin(), mem.end(), cmp);
    int sz = mem.size();
    for (int i = (0), iend = (sz - 1); i <= iend; i++) {
      int lca = LCA(mem[i], mem[(i + 1) % sz]);
      if (!inbuc[lca]) inbuc[lca] = 1, mem.push_back(lca);
    }
    sort(mem.begin(), mem.end(), cmp);
    int rt = build();
    ans = 0;
    dfs2(rt, 0);
    clear();
    printf("%lld\n", ans);
  }
  return 0;
}