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>
const int N = 400000 + 10;
using u64 = unsigned long long int;
int read() {
  int x = 0;
  char c;
  do {
    c = getchar();
  } while (!isdigit(c));
  do {
    x = x * 10 + c - '0';
    c = getchar();
  } while (isdigit(c));
  return x;
}
void readS(char *S) {
  int p = 0;
  char c;
  do {
    c = getchar();
  } while (!isalpha(c));
  do {
    S[++p] = c;
    c = getchar();
  } while (isalpha(c));
}
int par[N], len[N], tr[N][26], cnt, root, last;
std::vector<int> son[N];
int newnode(int l) {
  int o = ++cnt;
  len[o] = l;
  return o;
}
void init() { root = last = ++cnt; }
int clone(int l, int old) {
  int o = ++cnt;
  len[o] = l;
  par[o] = par[old];
  for (int i = 0; i < 26; i++) tr[o][i] = tr[old][i];
  return o;
}
int dep[N], sz[N], hs[N];
int ch_head[N], dfn[N], idx;
void dfs1(int u) {
  dep[u] = dep[par[u]] + 1;
  sz[u] = 1;
  for (int v : son[u]) {
    dfs1(v);
    sz[u] += sz[v];
    if (sz[v] > sz[hs[u]]) hs[u] = v;
  }
}
void dfs2(int u, int f) {
  dfn[u] = ++idx;
  ch_head[u] = f;
  if (hs[u]) dfs2(hs[u], f);
  for (int v : son[u])
    if (dfn[v] == 0) dfs2(v, v);
}
int lca(int x, int y) {
  while (ch_head[x] != ch_head[y]) {
    int tx = ch_head[x], ty = ch_head[y];
    if (dep[tx] < dep[ty])
      y = par[ty];
    else
      x = par[tx];
  }
  if (dep[x] < dep[y]) std::swap(x, y);
  return y;
}
void build() {
  for (int i = 1; i <= cnt; i++) son[par[i]].push_back(i);
  dfs1(root);
  dfs2(root, root);
}
void append(int c) {
  int np = newnode(len[last] + 1), p = last;
  while (p && tr[p][c] == 0) {
    tr[p][c] = np;
    p = par[p];
  }
  if (p == 0)
    par[np] = root;
  else {
    int q = tr[p][c];
    if (len[q] == len[p] + 1)
      par[np] = q;
    else {
      int nq = clone(len[p] + 1, q);
      par[q] = par[np] = nq;
      while (p && tr[p][c] == q) {
        tr[p][c] = nq;
        p = par[p];
      }
    }
  }
  last = np;
}
int n, m, at[N];
char S[N];
std::vector<int> vt[N];
int ca[N], cb[N], tag[N];
u64 ans = 0;
void solve(int u) {
  ca[u] = cb[u] = 0;
  if (tag[u] == 1) ca[u]++;
  if (tag[u] == 2) cb[u]++;
  if (tag[u] == 3) {
    ca[u]++;
    cb[u]++;
  }
  u64 cnt = ca[u] * cb[u];
  for (int v : vt[u]) {
    solve(v);
    cnt += (1ULL * ca[u] * cb[v] + 1ULL * cb[u] * ca[v]);
    ca[u] += ca[v];
    cb[u] += cb[v];
  }
  ans += cnt * len[u];
  tag[u] = 0;
}
int stk[N], top;
int main() {
  n = read();
  m = read();
  readS(S);
  init();
  for (int i = 1; i <= n; i++) {
    append(S[n - i + 1] - 'a');
    at[i] = last;
  }
  build();
  while (m--) {
    int A = read(), B = read();
    std::vector<int> v;
    for (int i = 0; i < A; i++) {
      int p = at[n - read() + 1];
      tag[p] += 1;
      v.push_back(p);
    }
    for (int i = 0; i < B; i++) {
      int p = at[n - read() + 1];
      tag[p] += 2;
      v.push_back(p);
    }
    std::sort(v.begin(), v.end(), [](int a, int b) { return dfn[a] < dfn[b]; });
    auto end = std::unique(v.begin(), v.end());
    top = 0;
    vt[stk[++top] = 1].clear();
    ca[1] = cb[1] = 0;
    for (auto i = v.begin(); i != end; i++) {
      int vtx = *i;
      if (vtx == 1) continue;
      int l = lca(vtx, stk[top]);
      if (l != stk[top]) {
        while (dfn[l] < dfn[stk[top - 1]]) {
          vt[stk[top - 1]].push_back(stk[top]);
          top--;
        }
        if (dfn[l] > dfn[stk[top - 1]]) {
          vt[l].clear();
          vt[l].push_back(stk[top]);
          stk[top] = l;
        } else
          vt[l].push_back(stk[top--]);
      }
      vt[vtx].clear();
      stk[++top] = vtx;
    }
    for (int i = 1; i < top; i++) vt[stk[i]].push_back(stk[i + 1]);
    ans = 0;
    solve(1);
    printf("%llu\n", ans);
  }
  return 0;
}