#include <bits/stdc++.h>
using namespace std;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, -1, 0, 1, -1, 0, 1};
long long a[100100];
map<int, long long> mp;
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
    mp[a[i]]++;
  }
  sort(a, a + n);
  if (mp[a[0]] > 2)
    cout << (mp[a[0]] - 2) * (mp[a[0]] - 1) * mp[a[0]] / 6 << endl;
  else if (mp[a[0]] == 2)
    cout << mp[a[2]] << endl;
  else if (mp[a[1]] > 1)
    cout << mp[a[1]] * (mp[a[1]] - 1) / 2 << endl;
  else
    cout << mp[a[2]] << endl;
  return 0;
}

### 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;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, -1, 0, 1, -1, 0, 1};
long long a[100100];
map<int, long long> mp;
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < (n); i++) {
    cin >> a[i];
    mp[a[i]]++;
  }
  sort(a, a + n);
  if (mp[a[0]] > 2)
    cout << (mp[a[0]] - 2) * (mp[a[0]] - 1) * mp[a[0]] / 6 << endl;
  else if (mp[a[0]] == 2)
    cout << mp[a[2]] << endl;
  else if (mp[a[1]] > 1)
    cout << mp[a[1]] * (mp[a[1]] - 1) / 2 << endl;
  else
    cout << mp[a[2]] << endl;
  return 0;
}

```