#include <bits/stdc++.h>
using namespace std;
long long C(long long n, long long k) {
  if (k == 3)
    return n * (n - 1) * (n - 2) / 6;
  else
    return n * (n - 1) / 2;
}
long long n, a[100005], ans, e1, e2, e3;
int main() {
  scanf("%lld", &n);
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  sort(a, a + n);
  e1 = upper_bound(a, a + n, a[0]) - a;
  if (e1 == 1) {
    e2 = upper_bound(a, a + n, a[1]) - a - 1;
    if (e2 == 1) {
      e3 = upper_bound(a, a + n, a[2]) - a - 2;
      ans = e3;
    } else
      ans = C(e2, 2);
  } else if (e1 == 2) {
    e2 = upper_bound(a, a + n, a[2]) - a - 2;
    ans = e2;
  } else
    ans = C(e1, 3);
  printf("%lld\n", 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;
long long C(long long n, long long k) {
  if (k == 3)
    return n * (n - 1) * (n - 2) / 6;
  else
    return n * (n - 1) / 2;
}
long long n, a[100005], ans, e1, e2, e3;
int main() {
  scanf("%lld", &n);
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  sort(a, a + n);
  e1 = upper_bound(a, a + n, a[0]) - a;
  if (e1 == 1) {
    e2 = upper_bound(a, a + n, a[1]) - a - 1;
    if (e2 == 1) {
      e3 = upper_bound(a, a + n, a[2]) - a - 2;
      ans = e3;
    } else
      ans = C(e2, 2);
  } else if (e1 == 2) {
    e2 = upper_bound(a, a + n, a[2]) - a - 2;
    ans = e2;
  } else
    ans = C(e1, 3);
  printf("%lld\n", ans);
  return 0;
}

```