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;
template <typename T>
void read(T &x) {
  x = 0;
  char ch = getchar();
  bool f = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = 1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 1) + (x << 3) + ch - '0', ch = getchar();
  x = f ? -x : x;
}
const int N = 2e5 + 7, M = 4e5 + 7;
int n, m, edp[N], pre[M];
char s[N];
int head[M], to[M], nxt[M], tot;
void adde(int u, int v) {
  to[++tot] = v;
  nxt[tot] = head[u];
  head[u] = tot;
}
struct SAM {
  const static int N = 4e5 + 7;
  int len[N], ch[N][26], fail[N], tot = 1, lst = 1;
  int ins(int c) {
    int now = ++tot, p = lst;
    len[now] = len[lst] + 1;
    lst = now;
    while (p && !ch[p][c]) ch[p][c] = now, p = fail[p];
    if (!p)
      fail[now] = 1;
    else {
      int q = ch[p][c];
      if (len[q] == len[p] + 1)
        fail[now] = q;
      else {
        int clo = ++tot;
        len[clo] = len[p] + 1;
        fail[clo] = fail[q];
        fail[q] = fail[now] = clo;
        memcpy(ch[clo], ch[q], sizeof(ch[q]));
        while (p && ch[p][c] == q) ch[p][c] = clo, p = fail[p];
      }
    }
    return now;
  }
  void build() {
    for (int i = (2), _ = (tot); i <= _; ++i) adde(fail[i], i);
  }
} sam;
struct SEG {
  const static int N = 4e5 + 7;
  long long sum[N << 2];
  int add[N << 2];
  void modify(int k, int l, int r, int x, int y, int v) {
    if (x <= l && r <= y) {
      add[k] += v;
      return;
    }
    int mid = (l + r) >> 1;
    sum[k] += (pre[min(r, y)] - pre[max(l, x) - 1]) * v;
    if (x <= mid) modify((k << 1), l, mid, x, y, v);
    if (mid < y) modify((k << 1 | 1), mid + 1, r, x, y, v);
  }
  long long query(int k, int l, int r, int x, int y) {
    if (x <= l && r <= y)
      return sum[k] + (long long)add[k] * (pre[r] - pre[l - 1]);
    int mid = (l + r) >> 1;
    long long ans = (long long)add[k] * (pre[min(r, y)] - pre[max(l, x) - 1]);
    if (x <= mid) ans += query((k << 1), l, mid, x, y);
    if (mid < y) ans += query((k << 1 | 1), mid + 1, r, x, y);
    return ans;
  }
} T;
struct SP {
  const static int N = 4e5 + 7;
  int siz[N], f[N], top[N], seg[N], son[N], dep[N], rev[N];
  void df(int u, int fa) {
    f[u] = fa;
    dep[u] = dep[fa] + 1;
    siz[u] = 1;
    for (int i = head[u], v = to[i]; i; i = nxt[i], v = to[i]) {
      df(v, u);
      siz[u] += siz[v];
      if (siz[v] > siz[son[u]]) son[u] = v;
    }
  }
  void dfs(int u) {
    seg[u] = ++seg[0];
    rev[seg[0]] = u;
    if (son[u]) {
      top[son[u]] = top[u];
      dfs(son[u]);
    } else
      return;
    for (int i = head[u], v = to[i]; i; i = nxt[i], v = to[i]) {
      if (!top[v]) {
        top[v] = v;
        dfs(v);
      }
    }
  }
  void init() {
    df(1, 0);
    top[1] = 1;
    dfs(1);
    for (int i = (1), _ = (sam.tot); i <= _; ++i)
      pre[i] = pre[i - 1] + sam.len[rev[i]] - sam.len[sam.fail[rev[i]]];
  }
  void upd(int x, int v) {
    while (x) T.modify(1, 1, sam.tot, seg[top[x]], seg[x], v), x = f[top[x]];
  }
  long long qry(int x) {
    long long ans = 0;
    while (x) ans += T.query(1, 1, sam.tot, seg[top[x]], seg[x]), x = f[top[x]];
    return ans;
  }
} sp;
signed main() {
  read(n);
  read(m);
  scanf("%s", s + 1);
  reverse(s + 1, s + n + 1);
  for (int i = (1), _ = (n); i <= _; ++i) edp[i] = sam.ins(s[i] - 'a');
  sam.build();
  sp.init();
  while (m--) {
    int x, y;
    read(x);
    read(y);
    long long ans = 0;
    vector<int> wt;
    while (x--) {
      int tp;
      read(tp);
      tp = n + 1 - tp;
      sp.upd(edp[tp], 1);
      wt.push_back(edp[tp]);
    }
    while (y--) {
      int tp;
      read(tp);
      tp = n + 1 - tp;
      ans += sp.qry(edp[tp]);
    }
    printf("%lld\n", ans);
    for (auto v : wt) sp.upd(v, -1);
  }
  return 0;
}