#include <bits/stdc++.h>
using namespace std;
vector<long long int> a;
long long int p = 0, e = 0;
int main() {
  ios ::sync_with_stdio(false);
  cin.tie(0);
  ;
  long long int i, l, j, k, n, m, t, x, y, f, d = 0, p = 0, q = 0, r = 0;
  cin >> n;
  a.resize(n);
  for (i = 0; i < n; i++) cin >> a[i];
  ;
  sort(a.begin(), a.end());
  for (i = 0; i <= n - 2; i++) {
    if (d == 0) {
      if (a[i] == a[i + 1]) {
        p++;
        if (i + 1 == n - 1) p++;
      } else {
        p++;
        d++;
        if (i + 1 == n - 1) q++;
      }
    } else if (d == 1) {
      if (a[i] == a[i + 1]) {
        q++;
        if (i + 1 == n - 1) q++;
      } else {
        q++;
        d++;
        if (i + 1 == n - 1) r++;
      }
    } else if (d == 2) {
      if (a[i] == a[i + 1]) {
        r++;
        if (i + 1 == n - 1) r++;
      } else {
        r++;
        d++;
      }
    } else
      break;
  }
  if (p >= 3) {
    d = p * (p - 1) * (p - 2);
    d /= 2;
    d /= 3;
    cout << d << endl;
  } else if (p == 2)
    cout << q << endl;
  else if (p == 1) {
    if (q >= 2) {
      d = q * (q - 1);
      d /= 2;
      cout << d << endl;
    } else
      cout << r << 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;
vector<long long int> a;
long long int p = 0, e = 0;
int main() {
  ios ::sync_with_stdio(false);
  cin.tie(0);
  ;
  long long int i, l, j, k, n, m, t, x, y, f, d = 0, p = 0, q = 0, r = 0;
  cin >> n;
  a.resize(n);
  for (i = 0; i < n; i++) cin >> a[i];
  ;
  sort(a.begin(), a.end());
  for (i = 0; i <= n - 2; i++) {
    if (d == 0) {
      if (a[i] == a[i + 1]) {
        p++;
        if (i + 1 == n - 1) p++;
      } else {
        p++;
        d++;
        if (i + 1 == n - 1) q++;
      }
    } else if (d == 1) {
      if (a[i] == a[i + 1]) {
        q++;
        if (i + 1 == n - 1) q++;
      } else {
        q++;
        d++;
        if (i + 1 == n - 1) r++;
      }
    } else if (d == 2) {
      if (a[i] == a[i + 1]) {
        r++;
        if (i + 1 == n - 1) r++;
      } else {
        r++;
        d++;
      }
    } else
      break;
  }
  if (p >= 3) {
    d = p * (p - 1) * (p - 2);
    d /= 2;
    d /= 3;
    cout << d << endl;
  } else if (p == 2)
    cout << q << endl;
  else if (p == 1) {
    if (q >= 2) {
      d = q * (q - 1);
      d /= 2;
      cout << d << endl;
    } else
      cout << r << endl;
  }
  return 0;
}

```