#include <bits/stdc++.h>
using namespace std;
const int inf = 1000000000 + 7;
const int N = 1e6 + 5;
long long n, a[N], b[N], c, ans;
int main() {
  scanf("%I64d", &n);
  for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    if (a[i] > a[i - 1]) c++;
    b[c]++;
  }
  if (b[1] >= 3) {
    ans = b[1] * (b[1] - 1) * (b[1] - 2) / 6;
  } else if (b[2] + b[1] >= 3) {
    if (b[1] == 1) {
      ans = b[2] * (b[2] - 1) / 2;
    } else {
      ans = b[2];
    }
  } else {
    ans = b[3];
  }
  printf("%I64d", ans);
}

### Prompt

Your challenge is to write 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;
const int inf = 1000000000 + 7;
const int N = 1e6 + 5;
long long n, a[N], b[N], c, ans;
int main() {
  scanf("%I64d", &n);
  for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    if (a[i] > a[i - 1]) c++;
    b[c]++;
  }
  if (b[1] >= 3) {
    ans = b[1] * (b[1] - 1) * (b[1] - 2) / 6;
  } else if (b[2] + b[1] >= 3) {
    if (b[1] == 1) {
      ans = b[2] * (b[2] - 1) / 2;
    } else {
      ans = b[2];
    }
  } else {
    ans = b[3];
  }
  printf("%I64d", ans);
}

```