#include <bits/stdc++.h>
using namespace std;
const int N = 1010101;
const int LOGN = (int)log2(N) + 2;
int SQRTN = (int)sqrt(N) + 7;
const long long infi = 1ll << 60;
long long n, a, ans;
vector<long long> vec;
map<long long, long long> rep;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  cin >> n;
  for (int(i) = (0); (i) <= (int)(n - 1); ++(i)) {
    cin >> a;
    vec.push_back(a);
    ++rep[a];
  }
  sort(vec.begin(), vec.end());
  if (vec[0] == vec[2])
    ans = rep[vec[0]] * (rep[vec[0]] - 1) * (rep[vec[0]] - 2) / 6;
  else if ((vec[0] == vec[1] && vec[1] != vec[2]) ||
           (vec[1] == vec[2] && vec[0] != vec[2])) {
    if (rep[vec[0]] == 1)
      ans = rep[vec[2]] * (rep[vec[2]] - 1) / 2;
    else
      ans = rep[vec[2]];
  } else
    ans = rep[vec[2]];
  cout << ans << endl;
  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;
const int N = 1010101;
const int LOGN = (int)log2(N) + 2;
int SQRTN = (int)sqrt(N) + 7;
const long long infi = 1ll << 60;
long long n, a, ans;
vector<long long> vec;
map<long long, long long> rep;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  cin >> n;
  for (int(i) = (0); (i) <= (int)(n - 1); ++(i)) {
    cin >> a;
    vec.push_back(a);
    ++rep[a];
  }
  sort(vec.begin(), vec.end());
  if (vec[0] == vec[2])
    ans = rep[vec[0]] * (rep[vec[0]] - 1) * (rep[vec[0]] - 2) / 6;
  else if ((vec[0] == vec[1] && vec[1] != vec[2]) ||
           (vec[1] == vec[2] && vec[0] != vec[2])) {
    if (rep[vec[0]] == 1)
      ans = rep[vec[2]] * (rep[vec[2]] - 1) / 2;
    else
      ans = rep[vec[2]];
  } else
    ans = rep[vec[2]];
  cout << ans << endl;
  return 0;
}

```