#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  int arr[n];
  for (int i = 0; i < n; i++) {
    unsigned long long x;
    cin >> x;
    arr[i] = x;
  }
  sort(arr, arr + n);
  if (arr[0] == arr[1] && arr[1] == arr[2]) {
    unsigned long long x = 3;
    while (arr[x] == arr[2]) {
      if (x < (n - 1)) {
        x++;
      } else {
        x++;
        break;
      }
    }
    unsigned long long w = (((x) * (x - 2) * (x - 1)) / 6);
    cout << w;
  } else if (arr[1] == arr[2]) {
    unsigned long long x = 3;
    while (arr[x] == arr[2]) {
      if (x < (n - 1)) {
        x++;
      } else {
        x++;
        break;
      }
    }
    cout << ((x - 2) * (x - 1)) / 2;
    ;
  } else {
    int x = 3;
    while (arr[x] == arr[2]) {
      if (x < (n - 1)) {
        x++;
      } else {
        x++;
        break;
      }
    }
    cout << (x - 2);
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
int main() {
  int n;
  cin >> n;
  int arr[n];
  for (int i = 0; i < n; i++) {
    unsigned long long x;
    cin >> x;
    arr[i] = x;
  }
  sort(arr, arr + n);
  if (arr[0] == arr[1] && arr[1] == arr[2]) {
    unsigned long long x = 3;
    while (arr[x] == arr[2]) {
      if (x < (n - 1)) {
        x++;
      } else {
        x++;
        break;
      }
    }
    unsigned long long w = (((x) * (x - 2) * (x - 1)) / 6);
    cout << w;
  } else if (arr[1] == arr[2]) {
    unsigned long long x = 3;
    while (arr[x] == arr[2]) {
      if (x < (n - 1)) {
        x++;
      } else {
        x++;
        break;
      }
    }
    cout << ((x - 2) * (x - 1)) / 2;
    ;
  } else {
    int x = 3;
    while (arr[x] == arr[2]) {
      if (x < (n - 1)) {
        x++;
      } else {
        x++;
        break;
      }
    }
    cout << (x - 2);
  }
  return 0;
}

```