#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int a[N];
map<int, int> b, q;
int main() {
  int n;
  cin >> n;
  for (int x = 0; x < n; x++) {
    cin >> a[x];
    b[a[x]]++;
  }
  sort(a, a + n);
  long long sum = 0;
  for (int x = 0; x < n; x++) {
    b[a[x]]--;
    if (a[x] == a[1]) {
      sum += 1LL * q[a[0]] * b[a[2]];
    }
    q[a[x]]++;
  }
  cout << sum << endl;
}

### Prompt

Your challenge is to write 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;
const int N = 1e5 + 10;
int a[N];
map<int, int> b, q;
int main() {
  int n;
  cin >> n;
  for (int x = 0; x < n; x++) {
    cin >> a[x];
    b[a[x]]++;
  }
  sort(a, a + n);
  long long sum = 0;
  for (int x = 0; x < n; x++) {
    b[a[x]]--;
    if (a[x] == a[1]) {
      sum += 1LL * q[a[0]] * b[a[2]];
    }
    q[a[x]]++;
  }
  cout << sum << endl;
}

```