#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  long long int arr[n], min1 = INT_MAX;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    if (arr[i] < min1) min1 = arr[i];
  }
  long long int count = 0, min2 = INT_MAX;
  for (int i = 0; i < n; i++) {
    if (arr[i] == min1) count++;
    if (arr[i] < min2 && arr[i] != min1) min2 = arr[i];
  }
  if (count >= 3) {
    cout << count * (count - 1) * (count - 2) / 6;
  } else {
    long long int count2 = 0;
    if (count == 2) {
      for (int i = 0; i < n; i++) {
        if (arr[i] == min2) count2++;
      }
      cout << count2;
    }
    if (count == 1) {
      for (int i = 0; i < n; i++) {
        if (arr[i] == min2) count2++;
      }
      if (count2 == 1) {
        long long int count3 = 0, min3 = INT_MAX;
        for (int i = 0; i < n; i++) {
          if (arr[i] < min3 && arr[i] != min2 && arr[i] != min1) min3 = arr[i];
        }
        for (int i = 0; i < n; i++) {
          if (arr[i] == min3) count3++;
        }
        cout << count3;
      }
      if (count2 == 2) {
        cout << 1;
      }
      if (count2 > 2) {
        cout << count2 * (count2 - 1) / 2;
      }
    }
  }
}

### Prompt

Please formulate 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;
  long long int arr[n], min1 = INT_MAX;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    if (arr[i] < min1) min1 = arr[i];
  }
  long long int count = 0, min2 = INT_MAX;
  for (int i = 0; i < n; i++) {
    if (arr[i] == min1) count++;
    if (arr[i] < min2 && arr[i] != min1) min2 = arr[i];
  }
  if (count >= 3) {
    cout << count * (count - 1) * (count - 2) / 6;
  } else {
    long long int count2 = 0;
    if (count == 2) {
      for (int i = 0; i < n; i++) {
        if (arr[i] == min2) count2++;
      }
      cout << count2;
    }
    if (count == 1) {
      for (int i = 0; i < n; i++) {
        if (arr[i] == min2) count2++;
      }
      if (count2 == 1) {
        long long int count3 = 0, min3 = INT_MAX;
        for (int i = 0; i < n; i++) {
          if (arr[i] < min3 && arr[i] != min2 && arr[i] != min1) min3 = arr[i];
        }
        for (int i = 0; i < n; i++) {
          if (arr[i] == min3) count3++;
        }
        cout << count3;
      }
      if (count2 == 2) {
        cout << 1;
      }
      if (count2 > 2) {
        cout << count2 * (count2 - 1) / 2;
      }
    }
  }
}

```