#include <bits/stdc++.h>
using namespace std;
const long long N = 1e5 + 5;
long long n;
map<long long, long long> a;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long c;
    cin >> c;
    a[c]++;
  }
  auto fi = *a.begin();
  if (fi.second >= 3) {
    long long k = fi.second;
    cout << (k * (k - 1) * (k - 2)) / 6 << endl;
  } else if (fi.second == 2) {
    long long F = fi.second;
    long long S = (*next(a.begin())).second;
    cout << (F * (F - 1)) / 2 * S << endl;
  } else {
    long long F = fi.second;
    long long S = (*next(a.begin())).second;
    if (S >= 2) {
      cout << F * (S * (S - 1)) / 2 << endl;
    } else {
      long long T = (*next(next(a.begin()))).second;
      cout << F * S * T << endl;
    }
  }
  return 0;
}

### 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 long long N = 1e5 + 5;
long long n;
map<long long, long long> a;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 0; i < n; i++) {
    long long c;
    cin >> c;
    a[c]++;
  }
  auto fi = *a.begin();
  if (fi.second >= 3) {
    long long k = fi.second;
    cout << (k * (k - 1) * (k - 2)) / 6 << endl;
  } else if (fi.second == 2) {
    long long F = fi.second;
    long long S = (*next(a.begin())).second;
    cout << (F * (F - 1)) / 2 * S << endl;
  } else {
    long long F = fi.second;
    long long S = (*next(a.begin())).second;
    if (S >= 2) {
      cout << F * (S * (S - 1)) / 2 << endl;
    } else {
      long long T = (*next(next(a.begin()))).second;
      cout << F * S * T << endl;
    }
  }
  return 0;
}

```