#include <bits/stdc++.h>
using namespace std;
map<int, int> cont;
int ans[3];
int main() {
  int n, d;
  cin >> n;
  while (n--) {
    cin >> d;
    cont[d]++;
  }
  map<int, int>::iterator it = cont.begin();
  int sum = 0, loc = 0;
  while (it != cont.end()) {
    sum += it->second;
    ans[loc++] = it->second;
    if (sum >= 3) break;
    it++;
  }
  long long fans = 0;
  if (loc == 1) {
    fans = ans[0];
    fans *= (fans - 1) * (fans - 2);
    fans /= 6;
  } else if (loc == 2) {
    if (ans[0] == 1) {
      fans = ans[1];
      fans = fans * (fans - 1) / 2;
    } else {
      fans = ans[1];
    }
  } else {
    fans = ans[2];
  }
  cout << fans << endl;
  return 0;
}

### Prompt

Please formulate 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> cont;
int ans[3];
int main() {
  int n, d;
  cin >> n;
  while (n--) {
    cin >> d;
    cont[d]++;
  }
  map<int, int>::iterator it = cont.begin();
  int sum = 0, loc = 0;
  while (it != cont.end()) {
    sum += it->second;
    ans[loc++] = it->second;
    if (sum >= 3) break;
    it++;
  }
  long long fans = 0;
  if (loc == 1) {
    fans = ans[0];
    fans *= (fans - 1) * (fans - 2);
    fans /= 6;
  } else if (loc == 2) {
    if (ans[0] == 1) {
      fans = ans[1];
      fans = fans * (fans - 1) / 2;
    } else {
      fans = ans[1];
    }
  } else {
    fans = ans[2];
  }
  cout << fans << endl;
  return 0;
}

```