#include <bits/stdc++.h>
using namespace std;
long long n, a[100005], conn[100005], last, p = 0;
long long C(long long n, long long r) {
  long long sum = 1;
  for (long long i = 1; i <= r; i++) sum = sum * (n + 1 - i) / i;
  return sum;
}
int main() {
  scanf("%lld", &n);
  for (int i = 0; i < n; ++i) scanf("%lld", &a[i]);
  sort(a, a + n);
  int con = 1;
  a[n] = -1;
  last = a[0];
  for (int i = 1; i <= n; ++i) {
    if (a[i] != last) {
      conn[p++] = con;
      last = a[i];
      con = 1;
    } else
      ++con;
  }
  if (conn[0] >= 3)
    printf("%lld\n", C(conn[0], 3));
  else if (conn[0] + conn[1] >= 3)
    printf("%lld\n", C(conn[1], 3 - conn[0]));
  else
    printf("%lld\n", conn[2]);
  return 0;
}

### Prompt

Your task is to create 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 n, a[100005], conn[100005], last, p = 0;
long long C(long long n, long long r) {
  long long sum = 1;
  for (long long i = 1; i <= r; i++) sum = sum * (n + 1 - i) / i;
  return sum;
}
int main() {
  scanf("%lld", &n);
  for (int i = 0; i < n; ++i) scanf("%lld", &a[i]);
  sort(a, a + n);
  int con = 1;
  a[n] = -1;
  last = a[0];
  for (int i = 1; i <= n; ++i) {
    if (a[i] != last) {
      conn[p++] = con;
      last = a[i];
      con = 1;
    } else
      ++con;
  }
  if (conn[0] >= 3)
    printf("%lld\n", C(conn[0], 3));
  else if (conn[0] + conn[1] >= 3)
    printf("%lld\n", C(conn[1], 3 - conn[0]));
  else
    printf("%lld\n", conn[2]);
  return 0;
}

```