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 = 2e5 + 5;
struct SAM {
  int cnt, last, par[maxn << 1], len[maxn << 1], trans[maxn << 1][26], id[maxn];
  int newnode(int parent, int length) {
    par[++cnt] = parent;
    len[cnt] = length;
    for (int i = 0; i <= (int)25; ++i) trans[cnt][i] = 0;
    return cnt;
  }
  void ini() {
    cnt = 0;
    last = newnode(0, 0);
  }
  void extend(int c, int pos) {
    int p = last;
    int np = newnode(1, len[last] + 1);
    id[pos] = np;
    while (p != 0 && !trans[p][c]) {
      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 <= (int)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;
int n, q;
char s[maxn];
int head[maxn << 1], cnt;
struct Edge {
  int to, nex;
} e[maxn << 2];
void adde(int u, int v) {
  e[++cnt].nex = head[u];
  e[cnt].to = v;
  head[u] = cnt;
  e[++cnt].nex = head[v];
  e[cnt].to = u;
  head[v] = cnt;
}
int dfn[maxn << 1], sz[maxn << 1], dep[maxn << 1], top[maxn << 1], f[maxn << 1],
    son[maxn << 1], Time;
void dfs1(int u, int fa) {
  sz[u] = 1;
  dep[u] = dep[fa] + 1;
  f[u] = fa;
  for (int i = head[u]; i != -1; i = e[i].nex) {
    int v = e[i].to;
    if (v == fa) continue;
    dfs1(v, u);
    sz[u] += sz[v];
    if (sz[v] > sz[son[u]]) son[u] = v;
  }
}
void dfs2(int u, int tt) {
  dfn[u] = ++Time;
  top[u] = tt;
  if (!son[u]) return;
  dfs2(son[u], tt);
  for (int i = head[u]; i != -1; i = e[i].nex) {
    int v = e[i].to;
    if (v != f[u] && v != son[u]) dfs2(v, v);
  }
}
int lca(int u, int v) {
  while (top[u] != top[v]) {
    if (dep[top[u]] < dep[top[v]]) swap(u, v);
    u = f[top[u]];
  }
  if (dep[u] < dep[v]) swap(u, v);
  return v;
}
int a[maxn << 1], b[maxn << 1], color[maxn << 1];
vector<int> g[maxn << 1];
int Top, stak[maxn << 1];
int cmp(const int& a, const int& b) { return dfn[a] < dfn[b]; }
void ins(int x) {
  if (Top == 1) {
    stak[++Top] = x;
    return;
  }
  int LCA = lca(x, stak[Top]);
  if (LCA == stak[Top]) {
    stak[++Top] = x;
    return;
  }
  while (Top > 1 && dfn[stak[Top - 1]] >= dfn[LCA]) {
    g[stak[Top - 1]].push_back(stak[Top]);
    Top--;
  }
  if (LCA != stak[Top]) {
    g[LCA].push_back(stak[Top]);
    stak[Top] = LCA;
  }
  stak[++Top] = x;
}
long long suma[maxn << 1], sumb[maxn << 1], sum[maxn << 1];
void dp(int u) {
  if (color[u] == 1) suma[u]++;
  if (color[u] == 2) sumb[u]++;
  if (color[u] == 3) suma[u]++, sumb[u]++, sum[u] = 1;
  for (int i = 0; i <= (int)g[u].size() - 1; ++i) {
    int v = g[u][i];
    dp(v);
    sum[u] += suma[u] * sumb[v];
    sum[u] += sumb[u] * suma[v];
    suma[u] += suma[v];
    sumb[u] += sumb[v];
  }
}
long long solve(int u) {
  long long res = sum[u] * sam.len[u];
  for (int i = 0; i <= (int)g[u].size() - 1; ++i) {
    int v = g[u][i];
    res += solve(v);
  }
  g[u].clear();
  suma[u] = sumb[u] = sum[u] = color[u] = 0;
  return res;
}
int main() {
  cin >> n >> q;
  scanf("%s", s + 1);
  reverse(s + 1, s + n + 1);
  sam.ini();
  for (int i = 1; i <= (int)n; ++i) sam.extend(s[i] - 'a', i);
  memset(head, -1, sizeof(head));
  cnt = -1;
  for (int i = 2; i <= (int)sam.cnt; ++i) adde(i, sam.par[i]);
  dfs1(1, 0);
  dfs2(1, 1);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    vector<int> v;
    for (int i = 1; i <= (int)x; ++i) {
      scanf("%d", &a[i]);
      v.push_back(sam.id[n - a[i] + 1]);
      color[sam.id[n - a[i] + 1]] = 1;
    }
    for (int i = 1; i <= (int)y; ++i) {
      scanf("%d", &b[i]);
      v.push_back(sam.id[n - b[i] + 1]);
      color[sam.id[n - b[i] + 1]] += 2;
    }
    sort(v.begin(), v.end(), cmp);
    v.erase(unique(v.begin(), v.end()), v.end());
    stak[Top = 1] = 1;
    for (int i = 0; i <= (int)v.size() - 1; ++i) ins(v[i]);
    while (Top > 0) {
      g[stak[Top - 1]].push_back(stak[Top]);
      Top--;
    }
    dp(1);
    cout << solve(1) << endl;
  }
}