#include <bits/stdc++.h>
using namespace std;
long long C(long long k, long long n) {
  long long res = 1;
  for (int i = 1; i <= k; i++) {
    res = res * (n - i + 1) / i;
  }
  return res;
}
int main() {
  int n, counting = 1, dis = 1, count[4];
  long a[100001];
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  if (a[1] == a[2] && a[2] == a[3]) {
    for (int i = 2; i <= n; i++) {
      if (a[i] == a[1])
        counting++;
      else
        break;
    }
    cout << C(3, counting);
  }
  if (a[1] != a[2] && a[2] == a[3]) {
    for (int i = 3; i <= n; i++) {
      if (a[i] == a[2]) counting++;
    }
    cout << C(2, counting);
  }
  if (a[1] != a[2] && a[2] != a[3]) {
    for (int i = 4; i <= n; i++) {
      if (a[i] == a[3]) counting++;
    }
    cout << counting;
  }
  if (a[1] == a[2] && a[2] != a[3]) {
    for (int i = 4; i <= n; i++) {
      if (a[i] == a[3]) counting++;
    }
    cout << counting;
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
long long C(long long k, long long n) {
  long long res = 1;
  for (int i = 1; i <= k; i++) {
    res = res * (n - i + 1) / i;
  }
  return res;
}
int main() {
  int n, counting = 1, dis = 1, count[4];
  long a[100001];
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  if (a[1] == a[2] && a[2] == a[3]) {
    for (int i = 2; i <= n; i++) {
      if (a[i] == a[1])
        counting++;
      else
        break;
    }
    cout << C(3, counting);
  }
  if (a[1] != a[2] && a[2] == a[3]) {
    for (int i = 3; i <= n; i++) {
      if (a[i] == a[2]) counting++;
    }
    cout << C(2, counting);
  }
  if (a[1] != a[2] && a[2] != a[3]) {
    for (int i = 4; i <= n; i++) {
      if (a[i] == a[3]) counting++;
    }
    cout << counting;
  }
  if (a[1] == a[2] && a[2] != a[3]) {
    for (int i = 4; i <= n; i++) {
      if (a[i] == a[3]) counting++;
    }
    cout << counting;
  }
}

```