#include <bits/stdc++.h>
using namespace std;
int s[300005] = {0};
int n = 0;
long long ans;
int count(int k) {
  int cnt = 0;
  int i = 1;
  while (i <= n && s[i] <= k) {
    if (s[i] == k) {
      cnt++;
    }
    i++;
  }
  return cnt;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
  }
  sort(s + 1, s + n + 1);
  int a = s[1], b = s[2], c = s[3];
  long long cnta = count(a);
  long long cntb = count(b);
  long long cntc = count(c);
  if (a == b && b == c) {
    ans = cnta * (cnta - 1) * (cnta - 2) / 6;
  } else {
    if (a == b) {
      ans = cntc * cnta * (cnta - 1) / 2;
    } else if (b == c) {
      ans = cnta * cntb * (cntb - 1) / 2;
    } else {
      ans = cnta * cntb * cntc;
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Create a solution in cpp for 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;
int s[300005] = {0};
int n = 0;
long long ans;
int count(int k) {
  int cnt = 0;
  int i = 1;
  while (i <= n && s[i] <= k) {
    if (s[i] == k) {
      cnt++;
    }
    i++;
  }
  return cnt;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
  }
  sort(s + 1, s + n + 1);
  int a = s[1], b = s[2], c = s[3];
  long long cnta = count(a);
  long long cntb = count(b);
  long long cntc = count(c);
  if (a == b && b == c) {
    ans = cnta * (cnta - 1) * (cnta - 2) / 6;
  } else {
    if (a == b) {
      ans = cntc * cnta * (cnta - 1) / 2;
    } else if (b == c) {
      ans = cnta * cntb * (cntb - 1) / 2;
    } else {
      ans = cnta * cntb * cntc;
    }
  }
  cout << ans;
  return 0;
}

```