#include <bits/stdc++.h>
using namespace std;
unsigned long long int fact(int n) {
  if (n == 0 || n == 1)
    return 1;
  else
    return n * fact(n - 1);
}
int main() {
  long long int n;
  cin >> n;
  long long int arr[1000000];
  for (long long int i = 0; i < n; ++i) {
    cin >> arr[i];
  }
  long long int min1, min2, min3;
  sort(arr, arr + n);
  min1 = arr[0];
  min2 = arr[1];
  min3 = arr[2];
  if (min1 == min2 && min2 == min3) {
    long long int c = 0;
    for (long long int i = 0; i < n; ++i) {
      if (arr[i] == min1) c++;
    }
    unsigned long long int ans;
    ans = (((c * (c - 1)) / 2) * (c - 2)) / 3;
    cout << ans;
    return 0;
  } else if (min1 == min2 && min2 != min3) {
    long long int c = 1;
    for (long long int i = 3; i < n; ++i) {
      if (arr[i] == min3) c++;
    }
    cout << c;
  } else if (min2 == min3) {
    long long int c = 1;
    for (long long int i = 2; i < n; ++i) {
      if (arr[i] == min2) c++;
    }
    long long int ans;
    ans = (c * (c - 1)) / 2;
    cout << ans;
    return 0;
  } else {
    long long int c = 1;
    for (long long int i = 3; i < n; ++i) {
      if (arr[i] == min3) c++;
    }
    cout << c;
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
unsigned long long int fact(int n) {
  if (n == 0 || n == 1)
    return 1;
  else
    return n * fact(n - 1);
}
int main() {
  long long int n;
  cin >> n;
  long long int arr[1000000];
  for (long long int i = 0; i < n; ++i) {
    cin >> arr[i];
  }
  long long int min1, min2, min3;
  sort(arr, arr + n);
  min1 = arr[0];
  min2 = arr[1];
  min3 = arr[2];
  if (min1 == min2 && min2 == min3) {
    long long int c = 0;
    for (long long int i = 0; i < n; ++i) {
      if (arr[i] == min1) c++;
    }
    unsigned long long int ans;
    ans = (((c * (c - 1)) / 2) * (c - 2)) / 3;
    cout << ans;
    return 0;
  } else if (min1 == min2 && min2 != min3) {
    long long int c = 1;
    for (long long int i = 3; i < n; ++i) {
      if (arr[i] == min3) c++;
    }
    cout << c;
  } else if (min2 == min3) {
    long long int c = 1;
    for (long long int i = 2; i < n; ++i) {
      if (arr[i] == min2) c++;
    }
    long long int ans;
    ans = (c * (c - 1)) / 2;
    cout << ans;
    return 0;
  } else {
    long long int c = 1;
    for (long long int i = 3; i < n; ++i) {
      if (arr[i] == min3) c++;
    }
    cout << c;
  }
}

```