#include <bits/stdc++.h>
using namespace std;
long long int choose(long long int n, long long int k) {
  if (k == 0) return 1;
  return (n * choose(n - 1, k - 1)) / k;
}
int main() {
  int n;
  long long int a[100001];
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  long long int index1 = 1, index2 = 1, index3 = 1;
  index1 = (upper_bound(a, a + n, a[0]) - a);
  if (index1 != n) {
    index2 = (upper_bound(a, a + n, a[index1]) - a);
    if (index2 != n) {
      index3 = (upper_bound(a, a + n, a[index2]) - a);
      index3 -= index2;
    }
    index2 -= index1;
  }
  if (index1 >= 3) {
    cout << choose(index1, 3);
  } else if (index1 + index2 >= 3) {
    cout << choose(index2, 3 - index1);
  } else {
    cout << index3;
  }
  cout << endl;
  return 0;
}

### Prompt

Your task is to create a CPP solution 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;
long long int choose(long long int n, long long int k) {
  if (k == 0) return 1;
  return (n * choose(n - 1, k - 1)) / k;
}
int main() {
  int n;
  long long int a[100001];
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  long long int index1 = 1, index2 = 1, index3 = 1;
  index1 = (upper_bound(a, a + n, a[0]) - a);
  if (index1 != n) {
    index2 = (upper_bound(a, a + n, a[index1]) - a);
    if (index2 != n) {
      index3 = (upper_bound(a, a + n, a[index2]) - a);
      index3 -= index2;
    }
    index2 -= index1;
  }
  if (index1 >= 3) {
    cout << choose(index1, 3);
  } else if (index1 + index2 >= 3) {
    cout << choose(index2, 3 - index1);
  } else {
    cout << index3;
  }
  cout << endl;
  return 0;
}

```