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;
function<void(void)> ____ = []() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
};
const int MAXN = 2e5 + 7;
struct SegmentTree {
  bool clr[MAXN << 2];
  int l[MAXN << 2], r[MAXN << 2], num[MAXN << 2];
  long long sum[MAXN << 2];
  void pushup(int rt) {
    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
    num[rt] = num[rt << 1] + num[rt << 1 | 1];
  }
  void build(int L, int R, int rt = 1) {
    l[rt] = L;
    r[rt] = R;
    clr[rt] = false;
    num[rt] = sum[rt] = 0;
    if (L + 1 == R) return;
    int mid = (L + R) >> 1;
    build(L, mid, rt << 1);
    build(mid, R, rt << 1 | 1);
  }
  void pushdown(int rt) {
    if (!clr[rt]) return;
    clr[rt << 1] = clr[rt << 1 | 1] = true;
    sum[rt << 1] = sum[rt << 1 | 1] = 0;
    num[rt << 1] = num[rt << 1 | 1] = 0;
    clr[rt] = false;
  }
  void clear(int L, int R, int rt = 1) {
    if (L >= r[rt] or l[rt] >= R) return;
    if (L <= l[rt] and r[rt] <= R) {
      clr[rt] = true;
      sum[rt] = num[rt] = 0;
      return;
    }
    pushdown(rt);
    clear(L, R, rt << 1);
    clear(L, R, rt << 1 | 1);
    pushup(rt);
  }
  void add(int pos, int x, int rt = 1) {
    if (l[rt] + 1 == r[rt]) {
      num[rt] += x;
      sum[rt] += 1ll * x * pos;
      return;
    }
    pushdown(rt);
    int mid = (l[rt] + r[rt]) >> 1;
    if (pos < mid)
      add(pos, x, rt << 1);
    else
      add(pos, x, rt << 1 | 1);
    pushup(rt);
  }
  int qnum(int L, int R, int rt = 1) {
    if (L >= r[rt] or l[rt] >= R) return 0;
    if (L <= l[rt] and r[rt] <= R) return num[rt];
    pushdown(rt);
    return qnum(L, R, rt << 1) + qnum(L, R, rt << 1 | 1);
  }
  long long qsum(int L, int R, int rt = 1) {
    if (L >= r[rt] or l[rt] >= R) return 0;
    if (L <= l[rt] and r[rt] <= R) return sum[rt];
    pushdown(rt);
    return qsum(L, R, rt << 1) + qsum(L, R, rt << 1 | 1);
  }
} seg;
int n, q, c[MAXN], rk[MAXN], sa[MAXN], sec[MAXN], lcp[MAXN], st[MAXN][20];
char s[MAXN];
void getSa(int m = 256) {
  for (int i = 0; i <= m; i++) c[i] = 0;
  for (int i = 1; i <= n; i++) c[rk[i] = s[i]]++;
  for (int i = 1; i <= m; i++) c[i] += c[i - 1];
  for (int i = 1; i <= n; i++) sa[c[rk[i]]--] = i;
  for (int k = 1; k <= n; k <<= 1) {
    int p = 0;
    for (int i = n - k + 1; i <= n; i++) sec[++p] = i;
    for (int i = 1; i <= n; i++)
      if (sa[i] > k) sec[++p] = sa[i] - k;
    for (int i = 0; i <= m; i++) c[i] = 0;
    for (int i = 1; i <= n; i++) c[rk[sec[i]]]++;
    for (int i = 1; i <= m; i++) c[i] += c[i - 1];
    for (int i = n; i >= 1; i--) sa[c[rk[sec[i]]]--] = sec[i];
    p = 0;
    swap(rk, sec);
    rk[sa[1]] = ++p;
    for (int i = 2; i <= n; i++)
      rk[sa[i]] =
          sec[sa[i]] == sec[sa[i - 1]] and sec[sa[i] + k] == sec[sa[i - 1] + k]
              ? p
              : ++p;
    if (p == n) break;
    m = p;
  }
}
void getHeight() {
  int k = 0;
  for (int i = 1; i <= n; i++) {
    if (k) k--;
    int j = sa[rk[i] - 1];
    while (s[i + k] == s[j + k]) k++;
    lcp[rk[i]] = k;
  }
}
void buildSt() {
  for (int i = 1; i <= n; i++) st[i][0] = lcp[i];
  for (int j = 1; (1 << j) <= n; j++)
    for (int i = 1; i + (1 << j) - 1 <= n; i++)
      st[i][j] = min(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
}
int qmin(int l, int r) {
  if (l > r) return 0;
  int d = (int)log2(r - l + 1);
  return min(st[l][d], st[r - (1 << d) + 1][d]);
}
int getLcp(int pa, int pb) {
  if (pa == pb) return n - pa + 1;
  int l = rk[pa], r = rk[pb];
  if (l > r) swap(l, r);
  return qmin(l + 1, r);
}
int sza, szb;
vector<int> veca, vecb;
void solveCase() {
  scanf("%d %d", &sza, &szb);
  veca.resize(sza);
  vecb.resize(szb);
  for (int &x : veca) scanf("%d", &x);
  for (int &x : vecb) scanf("%d", &x);
  sort(veca.begin(), veca.end(),
       [&](const int &a, const int &b) { return rk[a] < rk[b]; });
  sort(vecb.begin(), vecb.end(),
       [&](const int &a, const int &b) { return rk[a] < rk[b]; });
  long long ret = 0;
  seg.clear(0, n + 1);
  for (int i = 0, last = 0, ptr = 0; i < sza; last = rk[veca[i]], i++) {
    int rngmin = qmin(last + 1, rk[veca[i]]);
    int tot = seg.qnum(rngmin, n + 1);
    seg.clear(rngmin, n + 1);
    if (rngmin) seg.add(rngmin, tot);
    while (ptr < szb and rk[vecb[ptr]] <= rk[veca[i]]) {
      seg.add(getLcp(veca[i], vecb[ptr]), 1);
      ptr++;
    }
    ret += seg.qsum(0, n + 1);
  }
  seg.clear(0, n + 1);
  for (int i = sza - 1, last = n + 1, ptr = szb - 1; ~i;
       last = rk[veca[i]], i--) {
    int rngmin = qmin(rk[veca[i]] + 1, last);
    int tot = seg.qnum(rngmin, n + 1);
    seg.clear(rngmin, n + 1);
    if (rngmin) seg.add(rngmin, tot);
    while (~ptr and rk[vecb[ptr]] > rk[veca[i]]) {
      seg.add(getLcp(veca[i], vecb[ptr]), 1);
      ptr--;
    }
    ret += seg.qsum(0, n + 1);
  }
  cout << ret << endl;
}
void solve() {
  scanf("%d %d %s", &n, &q, s + 1);
  seg.build(0, n + 1);
  getSa();
  getHeight();
  buildSt();
  for (int kase = 1; kase <= q; kase++) solveCase();
}
int main() {
  solve();
  return 0;
}