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;
using ll = long long;
using pii = pair<int, int>;
using db = double;
using vi = vector<int>;
const int inf = 0x3f3f3f3f;
const db eps = 1e-8;
const int mod = 1e9 + 7;
ll qpow(ll a, ll b) {
  ll ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ret;
}
const int maxn = 400005;
int dep[maxn];
namespace SAM {
int ch[maxn][26], tot = 1, last = 1;
int fa[maxn], mx[maxn];
int ed[maxn];
void add(char x, int id) {
  x -= 'a';
  int p = last, np = ++tot;
  mx[np] = mx[p] + 1;
  ed[id] = np;
  for (; p && !ch[p][x]; p = fa[p]) ch[p][x] = np;
  if (!p)
    fa[np] = 1;
  else {
    int q = ch[p][x];
    if (mx[q] == mx[p] + 1)
      fa[np] = q;
    else {
      int nq = ++tot;
      memcpy(ch[nq], ch[q], sizeof(ch[nq]));
      mx[nq] = mx[p] + 1;
      fa[nq] = fa[q];
      fa[q] = fa[np] = nq;
      for (; p && ch[p][x] == q; p = fa[p]) ch[p][x] = nq;
    }
  }
  last = np;
}
int dfn[maxn], idx;
int p[maxn][20];
vi son[maxn];
void dfs(int x) {
  dfn[x] = ++idx;
  for (int i = 0; i < son[x].size(); ++i) {
    int u = son[x][i];
    dep[u] = dep[x] + 1;
    p[u][0] = x;
    for (int j = 1; j < 20; ++j) p[u][j] = p[p[u][j - 1]][j - 1];
    dfs(u);
  }
}
int lca(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  int d = dep[x] - dep[y];
  for (int i = 0; i < 20; ++i)
    if (d & (1 << i)) x = p[x][i];
  if (x == y) return x;
  for (int i = 19; i >= 0; --i) {
    if (p[x][i] != p[y][i]) {
      x = p[x][i];
      y = p[y][i];
    }
  }
  return p[x][0];
}
int q[maxn], cnt[maxn];
void build(int n) {
  for (int i = 1; i <= tot; ++i) cnt[mx[i]]++;
  for (int i = 1; i <= n; ++i) cnt[i] += cnt[i - 1];
  for (int i = 1; i <= tot; ++i) q[cnt[mx[i]]--] = i;
  for (int i = tot; i > 1; --i) {
    int x = q[i];
    son[fa[x]].push_back(x);
  }
  dep[1] = 1;
  dfs(1);
}
}  // namespace SAM
int a[maxn >> 1], b[maxn >> 1];
namespace VT {
int st[400005], top;
int sz1[400005], sz2[400005];
int num[400005], now[400005], p[400005], cnt;
bool cmp(int x, int y) { return SAM::dfn[x] < SAM::dfn[y]; }
void addEdge(int x, int y) {
  p[y] = x;
  now[++cnt] = y;
}
ll solve(int n, int m) {
  int tot = 0;
  for (int i = 1; i <= n; i++) {
    int x = SAM::ed[a[i]];
    num[++tot] = x;
    sz1[x]++;
  }
  for (int i = 1; i <= m; i++) {
    int x = SAM::ed[b[i]];
    num[++tot] = x;
    sz2[x]++;
  }
  sort(num + 1, num + tot + 1, cmp);
  int top = 0;
  st[++top] = 1;
  cnt = 0;
  for (int i = 1; i <= tot; i++) {
    int x = num[i];
    if (x == st[top]) continue;
    int p = SAM::lca(x, st[top]);
    for (;;) {
      if (dep[p] <= dep[st[top - 1]]) {
        addEdge(st[top - 1], st[top]);
        top--;
      } else if (dep[p] < dep[st[top]]) {
        addEdge(p, st[top]);
        st[top] = p;
        break;
      } else
        break;
    }
    if (p != st[top]) st[++top] = p;
    st[++top] = x;
  }
  for (; top; top--) addEdge(st[top - 1], st[top]);
  ll ans = 0;
  for (int i = 1; i <= cnt; i++) {
    int x = now[i];
    ans += (ll)sz1[x] * sz2[x] * (SAM::mx[x] - SAM::mx[p[x]]);
    sz1[p[x]] += sz1[x];
    sz2[p[x]] += sz2[x];
  }
  for (int i = 1; i <= cnt; i++) {
    int x = now[i];
    p[x] = sz1[x] = sz2[x] = 0;
  }
  return ans;
}
}  // namespace VT
char s[maxn >> 1];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  cin >> s + 1;
  for (int i = n; i > 0; --i) SAM::add(s[i], i);
  SAM::build(n);
  for (int i = 1; i <= m; ++i) {
    int x, y;
    cin >> x >> y;
    for (int j = 1; j <= x; ++j) cin >> a[j];
    for (int j = 1; j <= y; ++j) cin >> b[j];
    cout << VT::solve(x, y) << "\n";
  }
  return 0;
}