#include <bits/stdc++.h>
using namespace std;
map<int, int> m;
map<int, int>::iterator it;
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    m[x]++;
  }
  it = m.begin();
  int m1 = it->second;
  long long ans;
  if (m1 >= 3)
    ans = (1ll * m1 * (m1 - 1) * (m1 - 2)) / 6;
  else {
    it++;
    int m2 = it->second;
    if (m1 == 2)
      ans = m2;
    else if (m1 == 1) {
      if (m2 >= 2)
        ans = 1ll * m2 * (m2 - 1) / 2;
      else {
        it++;
        int m3 = it->second;
        ans = m3;
      }
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Generate 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;
map<int, int> m;
map<int, int>::iterator it;
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    m[x]++;
  }
  it = m.begin();
  int m1 = it->second;
  long long ans;
  if (m1 >= 3)
    ans = (1ll * m1 * (m1 - 1) * (m1 - 2)) / 6;
  else {
    it++;
    int m2 = it->second;
    if (m1 == 2)
      ans = m2;
    else if (m1 == 1) {
      if (m2 >= 2)
        ans = 1ll * m2 * (m2 - 1) / 2;
      else {
        it++;
        int m3 = it->second;
        ans = m3;
      }
    }
  }
  cout << ans;
  return 0;
}

```