#include <bits/stdc++.h>
using namespace std;
long long n, a[100100], rs;
map<long long, long long> M;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i], M[a[i]]++;
  sort(a + 1, a + n + 1);
  if (a[3] != a[2])
    cout << M[a[3]];
  else if (a[1] != a[2] && a[2] == a[3])
    cout << (1LL * M[a[3]] * (M[a[3]] - 1)) / 2LL;
  else if (a[1] == a[2] && a[2] == a[3]) {
    if (M[3] == 3) return cout << 1, 0;
    for (int i = 1; i <= M[a[3]] - 2; i++) rs += (1LL * i * (i + 1)) / 2LL;
    cout << rs;
  }
}

### Prompt

Your task is to create 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;
long long n, a[100100], rs;
map<long long, long long> M;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i], M[a[i]]++;
  sort(a + 1, a + n + 1);
  if (a[3] != a[2])
    cout << M[a[3]];
  else if (a[1] != a[2] && a[2] == a[3])
    cout << (1LL * M[a[3]] * (M[a[3]] - 1)) / 2LL;
  else if (a[1] == a[2] && a[2] == a[3]) {
    if (M[3] == 3) return cout << 1, 0;
    for (int i = 1; i <= M[a[3]] - 2; i++) rs += (1LL * i * (i + 1)) / 2LL;
    cout << rs;
  }
}

```