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>
const int inf = 0x3f3f3f3f, Inf = 0x7fffffff;
const long long INF = 0x7fffffffffffffff;
__inline__ __attribute__((always_inline)) unsigned int rnd() {
  static unsigned int seed = 416;
  return seed ^= seed >> 5, seed ^= seed << 17, seed ^= seed >> 13;
}
template <typename _Tp>
_Tp gcd(const _Tp &a, const _Tp &b) {
  return (!b) ? a : gcd(b, a % b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) _Tp abs(const _Tp &a) {
  return a >= 0 ? a : -a;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) _Tp max(const _Tp &a, const _Tp &b) {
  return a < b ? b : a;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) _Tp min(const _Tp &a, const _Tp &b) {
  return a < b ? a : b;
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void chmax(_Tp &a, const _Tp &b) {
  (a < b) && (a = b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void chmin(_Tp &a, const _Tp &b) {
  (b < a) && (a = b);
}
template <typename _Tp>
__inline__ __attribute__((always_inline)) void read(_Tp &x) {
  char ch(getchar());
  bool f(false);
  while (ch < 48 || ch > 57) f |= ch == 45, ch = getchar();
  x = ch & 15, ch = getchar();
  while (ch >= 48 && ch <= 57)
    x = (((x << 2) + x) << 1) + (ch & 15), ch = getchar();
  if (f) x = -x;
}
template <typename _Tp, typename... Args>
__inline__ __attribute__((always_inline)) void read(_Tp &t, Args &...args) {
  read(t);
  read(args...);
}
__inline__ __attribute__((always_inline)) int read_str(char *s) {
  char ch(getchar());
  while (ch == ' ' || ch == '\r' || ch == '\n') ch = getchar();
  char *tar = s;
  *tar = ch, ch = getchar();
  while (ch != ' ' && ch != '\r' && ch != '\n' && ch != EOF)
    *(++tar) = ch, ch = getchar();
  return tar - s + 1;
}
const int N = 200005;
int sa[N], rnk[N], tax[N], tp[N];
int height[N];
int n, m;
void Rsort() {
  memset(tax, 0, (m + 3) << 2);
  for (int i = 1; i <= n; ++i) ++tax[rnk[i]];
  for (int i = 1; i <= m; ++i) tax[i] += tax[i - 1];
  for (int i = n; i >= 1; --i) sa[tax[rnk[tp[i]]]--] = tp[i];
}
void SA(char *s) {
  for (int i = 1; i <= n; ++i) rnk[i] = s[i], tp[i] = i;
  m = 127;
  Rsort();
  for (int w = 1, p = 0; p < n; w <<= 1, m = p) {
    p = 0;
    for (int i = 0; i < w; ++i) tp[++p] = n - i;
    for (int i = 1; i <= n; ++i)
      if (sa[i] > w) tp[++p] = sa[i] - w;
    Rsort();
    std::swap(rnk, tp);
    rnk[sa[1]] = p = 1;
    for (int i = 2; i <= n; ++i)
      rnk[sa[i]] =
          (tp[sa[i - 1]] == tp[sa[i]] && tp[sa[i - 1] + w] == tp[sa[i] + w])
              ? p
              : ++p;
  }
  int k = 0;
  for (int i = 1; i <= n; ++i) {
    if (rnk[i] == 1) continue;
    if (k) --k;
    int j = sa[rnk[i] - 1];
    while (i + k <= n && j + k <= n && s[i + k] == s[j + k]) ++k;
    height[rnk[i]] = k;
  }
}
char s[N];
int f[N][21], Log2[N];
__inline__ __attribute__((always_inline)) int Query(int l, int r) {
  int k = Log2[r - l + 1];
  return min(f[l][k], f[r - (1 << k) + 1][k]);
}
std::pair<int, int> st[N];
int top;
long long solve(std::vector<int> v1, std::vector<int> v2) {
  top = 0;
  std::vector<std::pair<int, int> > v;
  for (auto it : v1) v.push_back(std::make_pair(it, 1));
  for (auto it : v2) v.push_back(std::make_pair(it, 0));
  std::sort(v.begin(), v.end());
  std::reverse(v.begin(), v.end());
  int last = 0;
  long long ans = 0, sum = 0;
  auto ins = [&](int qwq, int cnt) {
    while (top && qwq <= st[top].first) {
      sum -= 1ll * st[top].first * st[top].second;
      cnt += st[top].second;
      --top;
    }
    st[++top] = std::make_pair(qwq, cnt);
    sum += 1ll * qwq * cnt;
  };
  for (auto it : v) {
    if (last) {
      ins(Query(it.first + 1, last), 0);
    }
    if (it.second == 1) {
      ans += sum;
    } else {
      ins(height[it.first], 1);
    }
    last = it.first;
  }
  v.clear();
  return ans;
}
int main() {
  int _;
  read(n, _);
  read_str(s + 1);
  SA(s);
  for (int i = 2; i <= n; ++i) Log2[i] = Log2[i >> 1] + 1;
  for (int i = n; i >= 1; --i) {
    f[i][0] = height[i];
    for (int j = 1; j <= 20 && i + (1 << (j - 1)) <= n; ++j) {
      f[i][j] = min(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
    }
  }
  while (_--) {
    int a, b;
    read(a, b);
    std::vector<int> v1, v2;
    int x;
    for (int i = 1; i <= a; ++i) {
      read(x);
      v1.push_back(rnk[x]);
    }
    for (int i = 1; i <= b; ++i) {
      read(x);
      v2.push_back(rnk[x]);
    }
    std::sort(v1.begin(), v1.end());
    std::sort(v2.begin(), v2.end());
    long long ans = 0;
    for (auto it : v1) {
      auto itt = std::lower_bound(v2.begin(), v2.end(), it);
      if (itt != v2.end() && *itt == it) {
        ans += n - sa[it] + 1;
      }
    }
    printf("%lld\n", ans + solve(v1, v2) + solve(v2, v1));
  }
  return 0;
}