#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
int n;
long long a[N];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  int i = 0, j, k;
  while (i < n && a[i] == a[0]) i++;
  j = i;
  while (j < n && a[j] == a[1]) j++;
  k = j;
  while (k < n && a[k] == a[2]) k++;
  if (i >= 3) {
    long long n = i;
    cout << n * (n - 1ll) * (n - 2ll) / 6ll << endl;
  } else if (j >= 3) {
    long long n = j - i;
    if (i == 2)
      cout << n << endl;
    else
      cout << n * (n - 1ll) / 2ll;
  } else
    cout << k - j << endl;
}

### Prompt

Please create a solution in CPP 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;
const int N = 100005;
int n;
long long a[N];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  int i = 0, j, k;
  while (i < n && a[i] == a[0]) i++;
  j = i;
  while (j < n && a[j] == a[1]) j++;
  k = j;
  while (k < n && a[k] == a[2]) k++;
  if (i >= 3) {
    long long n = i;
    cout << n * (n - 1ll) * (n - 2ll) / 6ll << endl;
  } else if (j >= 3) {
    long long n = j - i;
    if (i == 2)
      cout << n << endl;
    else
      cout << n * (n - 1ll) / 2ll;
  } else
    cout << k - j << endl;
}

```