#include <bits/stdc++.h>
using namespace std;
const int MAXN = int(1e5);
const int MAXA = int(1e9);
int N;
int A[MAXN];
signed main() {
  ios::sync_with_stdio(0);
  cin >> N;
  map<int, int> cnt;
  for (int i = 0; i < N; ++i) {
    cin >> A[i];
    ++cnt[A[i]];
  }
  vector<int> m3;
  for (auto it = cnt.begin(); it != cnt.end(); ++it) {
    for (int i = 0; i < it->second and int(m3.size()) < 3; ++i) {
      m3.emplace_back(it->first);
    }
  }
  long long ans = 1;
  for (int i = 0; i < 3;) {
    int j = i;
    while (j < 3 and m3[j] == m3[i]) ++j;
    static auto C = [&](int n, int m) {
      if (m == 1) return 1LL * n;
      if (m == 2) return 1LL * n * (n - 1) / 2;
      return 1LL * n * (n - 1) * (n - 2) / 6;
    };
    ans *= C(cnt[m3[i]], j - i);
    i = j;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

After returning from the army Makes received a gift — an array a consisting of n positive integer numbers. He hadn't been solving problems for a long time, so he became interested to answer a particular question: how many triples of indices (i, j, k) (i < j < k), such that ai·aj·ak is minimum possible, are there in the array? Help him with it!

Input

The first line of input contains a positive integer number n (3 ≤ n ≤ 105) — the number of elements in array a. The second line contains n positive integer numbers ai (1 ≤ ai ≤ 109) — the elements of a given array.

Output

Print one number — the quantity of triples (i, j, k) such that i, j and k are pairwise distinct and ai·aj·ak is minimum possible.

Examples

Input

4
1 1 1 1


Output

4


Input

5
1 3 2 3 4


Output

2


Input

6
1 3 3 1 3 2


Output

1

Note

In the first example Makes always chooses three ones out of four, and the number of ways to choose them is 4.

In the second example a triple of numbers (1, 2, 3) is chosen (numbers, not indices). Since there are two ways to choose an element 3, then the answer is 2.

In the third example a triple of numbers (1, 1, 2) is chosen, and there's only one way to choose indices.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN = int(1e5);
const int MAXA = int(1e9);
int N;
int A[MAXN];
signed main() {
  ios::sync_with_stdio(0);
  cin >> N;
  map<int, int> cnt;
  for (int i = 0; i < N; ++i) {
    cin >> A[i];
    ++cnt[A[i]];
  }
  vector<int> m3;
  for (auto it = cnt.begin(); it != cnt.end(); ++it) {
    for (int i = 0; i < it->second and int(m3.size()) < 3; ++i) {
      m3.emplace_back(it->first);
    }
  }
  long long ans = 1;
  for (int i = 0; i < 3;) {
    int j = i;
    while (j < 3 and m3[j] == m3[i]) ++j;
    static auto C = [&](int n, int m) {
      if (m == 1) return 1LL * n;
      if (m == 2) return 1LL * n * (n - 1) / 2;
      return 1LL * n * (n - 1) * (n - 2) / 6;
    };
    ans *= C(cnt[m3[i]], j - i);
    i = j;
  }
  cout << ans << endl;
  return 0;
}

```