#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  int mins[3];
  int big = 2000000000;
  mins[0] = big;
  mins[1] = big;
  mins[2] = big;
  int a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == mins[0] || a[i] == mins[1]) continue;
    if (a[i] < mins[0]) {
      mins[2] = mins[1];
      mins[1] = mins[0];
      mins[0] = a[i];
    } else if (a[i] < mins[1]) {
      mins[2] = mins[1];
      mins[1] = a[i];
    } else if (a[i] < mins[2]) {
      mins[2] = a[i];
    }
  }
  long long n1 = 0, n2 = 0, n3 = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == mins[0]) n1++;
    if (a[i] == mins[1]) n2++;
    if (a[i] == mins[2]) n3++;
  }
  long long ans = 0;
  if (n1 >= 3) {
    ans = n1 * (n1 - 1) * (n1 - 2) / 6;
  } else if (n1 + n2 >= 3) {
    if (n1 == 2) {
      ans = n2;
    } else
      ans = n2 * (n2 - 1) / 2;
  } else
    ans = n1 * n2 * n3;
  cout << ans << endl;
}

### Prompt

Create a solution in CPP for 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 mins[3];
  int big = 2000000000;
  mins[0] = big;
  mins[1] = big;
  mins[2] = big;
  int a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == mins[0] || a[i] == mins[1]) continue;
    if (a[i] < mins[0]) {
      mins[2] = mins[1];
      mins[1] = mins[0];
      mins[0] = a[i];
    } else if (a[i] < mins[1]) {
      mins[2] = mins[1];
      mins[1] = a[i];
    } else if (a[i] < mins[2]) {
      mins[2] = a[i];
    }
  }
  long long n1 = 0, n2 = 0, n3 = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == mins[0]) n1++;
    if (a[i] == mins[1]) n2++;
    if (a[i] == mins[2]) n3++;
  }
  long long ans = 0;
  if (n1 >= 3) {
    ans = n1 * (n1 - 1) * (n1 - 2) / 6;
  } else if (n1 + n2 >= 3) {
    if (n1 == 2) {
      ans = n2;
    } else
      ans = n2 * (n2 - 1) / 2;
  } else
    ans = n1 * n2 * n3;
  cout << ans << endl;
}

```