#include <bits/stdc++.h>
using namespace std;
long long ncr(long long n, int r) {
  long long ans = 1;
  int i;
  for (i = 1; i <= r; i++) {
    ans *= n - r + i;
    ans /= i;
  }
  return ans;
}
int main() {
  int n, c[3], i, j, ind;
  long long x;
  vector<long long> a;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> x;
    a.push_back(x);
  }
  sort(a.begin(), a.end());
  for (i = 0; i < 3; i++) c[i] = 0;
  for (i = 0, j = 0, ind = 0; i < n; i++) {
    if (a[i] == a[ind])
      c[j]++;
    else {
      if (c[0] + c[1] + c[2] >= 3) break;
      j++;
      ind = i;
      i--;
    }
  }
  long long ans;
  if (c[2])
    ans = c[2];
  else if (c[1])
    ans = ncr(c[1], 3 - c[0]);
  else
    ans = ncr(c[0], 3);
  cout << ans;
  return 0;
}

### Prompt

Please create a solution in CPP 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 ncr(long long n, int r) {
  long long ans = 1;
  int i;
  for (i = 1; i <= r; i++) {
    ans *= n - r + i;
    ans /= i;
  }
  return ans;
}
int main() {
  int n, c[3], i, j, ind;
  long long x;
  vector<long long> a;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> x;
    a.push_back(x);
  }
  sort(a.begin(), a.end());
  for (i = 0; i < 3; i++) c[i] = 0;
  for (i = 0, j = 0, ind = 0; i < n; i++) {
    if (a[i] == a[ind])
      c[j]++;
    else {
      if (c[0] + c[1] + c[2] >= 3) break;
      j++;
      ind = i;
      i--;
    }
  }
  long long ans;
  if (c[2])
    ans = c[2];
  else if (c[1])
    ans = ncr(c[1], 3 - c[0]);
  else
    ans = ncr(c[0], 3);
  cout << ans;
  return 0;
}

```