#include <bits/stdc++.h>
using namespace std;
const int N = 4010;
long long a[100010];
map<long long, long long> mp;
signed main() {
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i], mp[a[i]]++;
  sort(a + 1, a + 1 + n);
  unique(a + 1, a + 1 + n);
  long long tt = mp[a[1]];
  if (tt >= 3) {
    cout << tt * (tt - 1) * (tt - 2) / 6 << endl;
  } else {
    long long tt2 = mp[a[2]];
    long long tt3 = mp[a[3]];
    if (tt == 2) {
      cout << tt2 << endl;
    } else {
      if (tt2 >= 2) {
        cout << tt2 * (tt2 - 1) / 2 << endl;
      } else
        cout << tt3 << endl;
    }
  }
}

### 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;
const int N = 4010;
long long a[100010];
map<long long, long long> mp;
signed main() {
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i], mp[a[i]]++;
  sort(a + 1, a + 1 + n);
  unique(a + 1, a + 1 + n);
  long long tt = mp[a[1]];
  if (tt >= 3) {
    cout << tt * (tt - 1) * (tt - 2) / 6 << endl;
  } else {
    long long tt2 = mp[a[2]];
    long long tt3 = mp[a[3]];
    if (tt == 2) {
      cout << tt2 << endl;
    } else {
      if (tt2 >= 2) {
        cout << tt2 * (tt2 - 1) / 2 << endl;
      } else
        cout << tt3 << endl;
    }
  }
}

```