#include <bits/stdc++.h>
using namespace std;
unsigned long long fact(unsigned long long n) {
  if (n == 0) return 1;
  return n * fact(n - 1);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  long long arr[n];
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
  }
  sort(arr, arr + n);
  long long count = 0;
  long long x = -1, y = -1, z = -1;
  long long c1 = 0, c2 = 0, c3 = 0, flag = 0;
  for (long long i = 0; i < n; i++) {
    if (x == -1) {
      x = arr[i];
    } else if (arr[i] > x && y == -1 && x != -1) {
      y = arr[i];
    } else if (arr[i] > y && z == -1 && y != -1) {
      z = arr[i];
    }
  }
  for (long long i = 0; i < n; i++) {
    if (arr[i] == x) {
      c1++;
    } else if (arr[i] == y) {
      c2++;
    } else if (arr[i] == z) {
      c3++;
    }
  }
  if (c1 == 3) {
    cout << 1 << endl;
  } else if (c1 > 3) {
    cout << ((c1 * (c1 - 1) * (c1 - 2)) / 6) << endl;
  } else {
    if (c1 == 1) {
      if (c2 == 1) {
        if (c3 == 1) {
          cout << 1 << endl;
        } else {
          cout << c3 << endl;
        }
      } else if (c2 > 1) {
        if (c2 == 2) {
          cout << 1 << endl;
        } else {
          cout << (c2 * (c2 - 1)) / (2) << endl;
        }
      }
    } else if (c1 == 2) {
      if (c2 == 1) {
        cout << 1 << endl;
      } else {
        cout << c2 << endl;
      }
    }
  }
}

### 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;
unsigned long long fact(unsigned long long n) {
  if (n == 0) return 1;
  return n * fact(n - 1);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  long long arr[n];
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
  }
  sort(arr, arr + n);
  long long count = 0;
  long long x = -1, y = -1, z = -1;
  long long c1 = 0, c2 = 0, c3 = 0, flag = 0;
  for (long long i = 0; i < n; i++) {
    if (x == -1) {
      x = arr[i];
    } else if (arr[i] > x && y == -1 && x != -1) {
      y = arr[i];
    } else if (arr[i] > y && z == -1 && y != -1) {
      z = arr[i];
    }
  }
  for (long long i = 0; i < n; i++) {
    if (arr[i] == x) {
      c1++;
    } else if (arr[i] == y) {
      c2++;
    } else if (arr[i] == z) {
      c3++;
    }
  }
  if (c1 == 3) {
    cout << 1 << endl;
  } else if (c1 > 3) {
    cout << ((c1 * (c1 - 1) * (c1 - 2)) / 6) << endl;
  } else {
    if (c1 == 1) {
      if (c2 == 1) {
        if (c3 == 1) {
          cout << 1 << endl;
        } else {
          cout << c3 << endl;
        }
      } else if (c2 > 1) {
        if (c2 == 2) {
          cout << 1 << endl;
        } else {
          cout << (c2 * (c2 - 1)) / (2) << endl;
        }
      }
    } else if (c1 == 2) {
      if (c2 == 1) {
        cout << 1 << endl;
      } else {
        cout << c2 << endl;
      }
    }
  }
}

```