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;
struct SAM {
  static const int V = 2e5 + 10;
  int cnt, last, par[V << 1], len[V << 1], id[V];
  int trans[V << 1][26];
  int newnode(int parent, int length) {
    par[++cnt] = parent;
    len[cnt] = length;
    for (int i = 0; i <= (25); ++i) trans[cnt][i] = -1;
    return cnt;
  }
  void ini() {
    cnt = 0;
    last = newnode(0, 0);
  }
  void extend(int pos, int c) {
    int p = last;
    int np = newnode(1, len[last] + 1);
    id[pos] = np;
    while (p != 0 && trans[p][c] == -1) {
      trans[p][c] = np;
      p = par[p];
    }
    if (p != 0) {
      int q = trans[p][c];
      if (len[q] == len[p] + 1)
        par[np] = q;
      else {
        int nq = newnode(par[q], len[p] + 1);
        for (int i = 0; i <= (25); ++i) trans[nq][i] = trans[q][i];
        par[np] = par[q] = nq;
        while (trans[p][c] == q) {
          trans[p][c] = nq;
          p = par[p];
        }
      }
    }
    last = np;
  }
} sam;
struct graph {
  static const int V = 4e5 + 5, E = 2 * V;
  int head[V];
  int to[E], nex[E], tot;
  inline void _addedge(int u, int v) {
    to[++tot] = v, nex[tot] = head[u], head[u] = tot;
  }
  inline void addedge(int u, int v) { _addedge(u, v), _addedge(v, u); }
  void deltree(int u, int father) {
    for (int i = head[u]; i; i = nex[i])
      if (to[i] != father) deltree(to[i], u);
    head[u] = 0;
  }
  int dep[V], dad[V], siz[V], son[V], dfn[V], chain[V], step;
  void dfs1(int u, int father) {
    siz[u] = 1;
    son[u] = 0;
    dad[u] = father;
    dep[u] = dep[father] + 1;
    for (int i = head[u]; i; i = nex[i]) {
      if (to[i] == father) continue;
      dfs1(to[i], u);
      siz[u] += siz[to[i]];
      if (siz[to[i]] > siz[son[u]]) son[u] = to[i];
    }
  }
  void dfs2(int u, int s) {
    dfn[u] = ++step;
    chain[u] = s;
    if (son[u]) dfs2(son[u], s);
    for (int i = head[u]; i; i = nex[i]) {
      if (to[i] == dad[u] || to[i] == son[u]) continue;
      dfs2(to[i], to[i]);
    }
  }
  int getlca(int x, int y) {
    while (chain[x] != chain[y]) {
      if (dep[chain[x]] < dep[chain[y]]) swap(x, y);
      x = dad[chain[x]];
    }
    return dep[x] < dep[y] ? x : y;
  }
  void buildvt(int* vert, int nums, graph& res) {
    static int *cmp, stk[V];
    cmp = dfn;
    sort(vert + 1, vert + nums + 1,
         [](int x, int y) { return cmp[x] < cmp[y]; });
    int top = 0;
    stk[++top] = 1;
    for (int i = vert[1] == 1 ? 2 : 1; i <= (nums); ++i) {
      int lca = getlca(vert[i], stk[top]);
      if (lca == stk[top]) {
        stk[++top] = vert[i];
        continue;
      }
      while (dfn[lca] <= dfn[stk[top - 1]])
        res.addedge(stk[top], stk[top - 1]), top--;
      if (lca != stk[top]) res.addedge(stk[top], lca), stk[top] = lca;
      stk[++top] = vert[i];
    }
    while (top >= 2) {
      res.addedge(stk[top], stk[top - 1]);
      top--;
    }
  }
  int color[V], sz1[V], sz2[V];
  long long solve(int u, int father) {
    long long res = 0;
    sz1[u] = 0, sz2[u] = 0;
    if (color[u] & 1) sz1[u]++;
    if (color[u] & 2) sz2[u]++;
    res += 1ll * sz1[u] * sz2[u] * sam.len[u];
    for (int i = head[u]; i; i = nex[i]) {
      if (to[i] == father) continue;
      res += solve(to[i], u) + 1ll * sz1[u] * sz2[to[i]] * sam.len[u] +
             1ll * sz2[u] * sz1[to[i]] * sam.len[u];
      sz1[u] += sz1[to[i]];
      sz2[u] += sz2[to[i]];
    }
    return res;
  }
} tree, vt;
long long query(vector<int>& vec) {
  tree.buildvt(vec.data() - 1, vec.size(), vt);
  long long res = vt.solve(1, 0);
  vt.deltree(1, 0);
  vt.tot = 0;
  return res;
}
inline int read() {
  int x;
  scanf("%d", &x);
  return x;
}
const int N = 2e5 + 5;
char s[N];
int main() {
  int len, q;
  scanf("%d%d%s", &len, &q, s + 1);
  reverse(s + 1, s + 1 + len);
  sam.ini();
  for (int i = 1; i <= (len); ++i) sam.extend(i, s[i] - 'a');
  for (int i = 2; i <= (sam.cnt); ++i) tree.addedge(i, sam.par[i]);
  tree.dfs1(1, 0), tree.dfs2(1, 1);
  for (int i = 1; i <= (q); ++i) {
    int n = read(), m = read();
    vector<int> c;
    for (int i = 1; i <= (n); ++i) {
      int x = sam.id[len + 1 - read()];
      c.push_back(x), vt.color[x] |= 1;
    }
    for (int i = 1; i <= (m); ++i) {
      int x = sam.id[len + 1 - read()];
      c.push_back(x), vt.color[x] |= 2;
    }
    sort(c.begin(), c.end());
    c.erase(unique(c.begin(), c.end()), c.end());
    printf("%lld\n", query(c));
    for (int x : c) vt.color[x] = 0;
  }
  tree.deltree(1, 0);
}