#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  int number[n];
  for (int i = 0; i < n; i++) cin >> number[i];
  sort(number, number + n);
  long long n0 = 0, n1 = 0, n2 = 0, com, com1;
  for (int i = 0; i < n; i++) {
    if (number[0] == number[i]) n0++;
  }
  if (n0 == 1) {
    for (int i = 1; i < n; i++) {
      if (number[1] == number[i]) n1++;
    }
  }
  if (n1 == 1 || n0 == 2) {
    for (int i = 2; i < n; i++) {
      if (number[2] == number[i]) n2++;
    }
  }
  if (n0 > 3) {
    com = n0 * (n0 - 1) * (n0 - 2);
    com /= 6;
    cout << com << endl;
    return 0;
  }
  if (n1 > 2) {
    com1 = n1 * (n1 - 1);
    com1 /= 2;
    cout << com1 << endl;
    return 0;
  }
  if (n2 > 1) {
    cout << n2 << endl;
    return 0;
  }
  cout << "1" << endl;
  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 number[n];
  for (int i = 0; i < n; i++) cin >> number[i];
  sort(number, number + n);
  long long n0 = 0, n1 = 0, n2 = 0, com, com1;
  for (int i = 0; i < n; i++) {
    if (number[0] == number[i]) n0++;
  }
  if (n0 == 1) {
    for (int i = 1; i < n; i++) {
      if (number[1] == number[i]) n1++;
    }
  }
  if (n1 == 1 || n0 == 2) {
    for (int i = 2; i < n; i++) {
      if (number[2] == number[i]) n2++;
    }
  }
  if (n0 > 3) {
    com = n0 * (n0 - 1) * (n0 - 2);
    com /= 6;
    cout << com << endl;
    return 0;
  }
  if (n1 > 2) {
    com1 = n1 * (n1 - 1);
    com1 /= 2;
    cout << com1 << endl;
    return 0;
  }
  if (n2 > 1) {
    cout << n2 << endl;
    return 0;
  }
  cout << "1" << endl;
  return 0;
}

```