#include <bits/stdc++.h>
using namespace std;
long long cnt[5];
long long comb(long long n, long long r) {
  if (n == r || r == 0) return 1;
  if (r == 2) return (n * (n - 1)) / 2;
  if (r == 3) return (n * (n - 1) * (n - 2)) / 6;
}
int main() {
  long long t, n, m, x, y, i;
  cin >> n;
  long long a[n];
  memset(cnt, 0, 5);
  for (i = 0; i <= n - 1; i++) cin >> a[i];
  sort(a, a + n);
  i = 0;
  while (a[i] == a[0] && i < n) {
    cnt[0]++;
    i++;
  }
  i = 1;
  while (a[i] == a[1] && i < n) {
    cnt[1]++;
    i++;
  }
  i = 2;
  while (a[i] == a[2] && i < n) {
    cnt[2]++;
    i++;
  }
  if (cnt[0] >= 3)
    x = comb(cnt[0], 3);
  else if (cnt[1] >= 2)
    x = comb(cnt[1], 2);
  else
    x = cnt[2];
  cout << x << endl;
  return 0;
}

### Prompt

Generate a cpp solution 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 cnt[5];
long long comb(long long n, long long r) {
  if (n == r || r == 0) return 1;
  if (r == 2) return (n * (n - 1)) / 2;
  if (r == 3) return (n * (n - 1) * (n - 2)) / 6;
}
int main() {
  long long t, n, m, x, y, i;
  cin >> n;
  long long a[n];
  memset(cnt, 0, 5);
  for (i = 0; i <= n - 1; i++) cin >> a[i];
  sort(a, a + n);
  i = 0;
  while (a[i] == a[0] && i < n) {
    cnt[0]++;
    i++;
  }
  i = 1;
  while (a[i] == a[1] && i < n) {
    cnt[1]++;
    i++;
  }
  i = 2;
  while (a[i] == a[2] && i < n) {
    cnt[2]++;
    i++;
  }
  if (cnt[0] >= 3)
    x = comb(cnt[0], 3);
  else if (cnt[1] >= 2)
    x = comb(cnt[1], 2);
  else
    x = cnt[2];
  cout << x << endl;
  return 0;
}

```