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;
inline int read() {
  char ch = getchar();
  int u = 0, f = 1;
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    u = (u << 1) + (u << 3) + ch - 48;
    ch = getchar();
  }
  return u * f;
}
const int maxn = 1e6 + 10;
vector<int> e[maxn];
int dep[maxn], size[maxn], top[maxn], coc = 0, dfn[maxn], up[maxn];
int is_val[maxn];
inline bool cmp(int x, int y) { return dfn[x] < dfn[y]; }
pair<long long, long long> rem[maxn];
vector<int> match;
class SAM {
 public:
  int par[maxn], mxl[maxn], last, cnt, ch[maxn][26];
  SAM() { cnt = last = 1; }
  inline void add(int c) {
    int p = last, np = ++cnt;
    last = np;
    mxl[np] = mxl[p] + 1;
    for (; p && !ch[p][c]; p = par[p]) ch[p][c] = np;
    if (!p)
      par[np] = 1;
    else {
      int q = ch[p][c];
      if (mxl[q] == mxl[p] + 1)
        par[np] = q;
      else {
        int nq = ++cnt;
        mxl[nq] = mxl[p] + 1;
        memcpy(ch[nq], ch[q], sizeof ch[nq]);
        par[nq] = par[q];
        par[q] = par[np] = nq;
        for (; p && ch[p][c] == q; p = par[p]) ch[p][c] = nq;
      }
    }
    match.push_back(np);
  }
  inline void Build_Suffix_Tree() {
    for (int i = 2; i <= cnt; ++i) e[par[i]].push_back(i);
  }
  inline void dfs(int x) {
    dep[x] = dep[par[x]] + 1;
    dfn[x] = ++coc;
    size[x] = 1;
    for (int i = 0; i < e[x].size(); ++i)
      dfs(e[x][i]), size[x] += size[e[x][i]];
  }
  inline void dfs(int x, int y) {
    top[x] = y;
    int pos = 0;
    for (int i = 0; i < e[x].size(); ++i)
      if (size[e[x][i]] > size[pos]) pos = e[x][i];
    if (!pos) return;
    dfs(pos, y);
    for (int i = 0; i < e[x].size(); ++i)
      if (e[x][i] != pos) dfs(e[x][i], e[x][i]);
  }
  inline int lca(int x, int y) {
    while (top[x] != top[y]) {
      if (dep[top[x]] < dep[top[y]]) swap(x, y);
      x = par[top[x]];
    }
    return dep[x] < dep[y] ? x : y;
  }
  inline vector<int> Build_Virtual_Tree(vector<int>& qry) {
    sort(qry.begin(), qry.end(), cmp);
    vector<int> Rtn = qry;
    stack<int> stk;
    for (int i = 0; i < qry.size(); ++i) {
      if (!stk.size())
        stk.push(qry[i]);
      else {
        int u = stk.top();
        int v = lca(qry[i], u);
        if (v == u)
          stk.push(qry[i]);
        else {
          while (1) {
            stk.pop();
            if (!stk.size() || dep[stk.top()] <= dep[v]) break;
            up[u] = stk.top();
            u = stk.top();
          }
          if (!stk.size() || stk.top() != v) {
            stk.push(v);
            Rtn.push_back(v);
          }
          up[u] = v;
          stk.push(qry[i]);
        }
      }
    }
    int u = stk.top();
    stk.pop();
    while (stk.size()) {
      up[u] = stk.top();
      u = stk.top();
      stk.pop();
    }
    sort(Rtn.begin(), Rtn.end(), cmp);
    return Rtn;
  }
} sam;
int n, m;
char s[maxn];
int main() {
  n = read(), m = read();
  scanf("%s", s + 1);
  reverse(s + 1, s + 1 + n);
  for (int i = 1; i <= n; ++i) sam.add(s[i] - 'a');
  sam.Build_Suffix_Tree();
  sam.dfs(1);
  sam.dfs(1, 1);
  for (int curtask = 1; curtask <= m; ++curtask) {
    int k1 = read(), k2 = read();
    vector<int> List;
    for (int j = 1; j <= k1 + k2; ++j) {
      int re = match[n - read()];
      is_val[re] |= ((j <= k1) ? 1 : 2);
      List.push_back(re);
    }
    sort(List.begin(), List.end());
    vector<int>::iterator It = unique(List.begin(), List.end());
    List.erase(It, List.end());
    List.push_back(1);
    vector<int> Nw = sam.Build_Virtual_Tree(List);
    long long ans = 0;
    for (int i = Nw.size() - 1; ~i; --i) {
      int curx = Nw[i];
      if (is_val[curx] == 1) rem[curx].first++;
      if (is_val[curx] == 2) rem[curx].second++;
      if (is_val[curx] == 3) rem[curx].first++, rem[curx].second++;
      ans += 1ll * (sam.mxl[curx] - sam.mxl[up[curx]]) *
             (rem[curx].first * rem[curx].second);
      rem[up[curx]].first += rem[curx].first;
      rem[up[curx]].second += rem[curx].second;
    }
    printf("%lld\n", ans);
    for (int j = 0; j < Nw.size(); ++j)
      rem[Nw[j]].first = rem[Nw[j]].second = 0;
    for (int j = 0; j < List.size(); ++j) is_val[List[j]] = 0;
    List.clear();
  }
  return 0;
}