#include <bits/stdc++.h>
using namespace std;
long long C(long long x, long long y) {
  long long ans = 1;
  for (long long i = 1; i <= y; i++) {
    ans *= x - i + 1;
    ans /= i;
  }
  return ans;
}
map<long long, long long> mp;
void solve() {
  long long n;
  cin >> n;
  for (int i = 0, a; i < n; i++) {
    cin >> a;
    mp[a]++;
  }
  map<long long, long long>::iterator it = mp.begin();
  long long x[3], ans;
  for (int i = 0; i < 3 && it != mp.end(); i++, it++) x[i] = it->second;
  if (x[0] >= 3)
    ans = C(x[0], 3);
  else if (x[0] + x[1] >= 3)
    ans = C(x[1], 3 - x[0]);
  else
    ans = C(x[2], 3 - x[0] - x[1]);
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  solve();
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 C(long long x, long long y) {
  long long ans = 1;
  for (long long i = 1; i <= y; i++) {
    ans *= x - i + 1;
    ans /= i;
  }
  return ans;
}
map<long long, long long> mp;
void solve() {
  long long n;
  cin >> n;
  for (int i = 0, a; i < n; i++) {
    cin >> a;
    mp[a]++;
  }
  map<long long, long long>::iterator it = mp.begin();
  long long x[3], ans;
  for (int i = 0; i < 3 && it != mp.end(); i++, it++) x[i] = it->second;
  if (x[0] >= 3)
    ans = C(x[0], 3);
  else if (x[0] + x[1] >= 3)
    ans = C(x[1], 3 - x[0]);
  else
    ans = C(x[2], 3 - x[0] - x[1]);
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  solve();
  return 0;
}

```