#include <bits/stdc++.h>
using namespace std;
const int maxn = 100000 + 10;
int a[maxn], b[3];
long long cal(int n, int m) {
  long long n1 = 1;
  long long m1 = 1;
  for (int i = n; i > n - m; i--) {
    n1 *= i;
    m1 *= (n - i + 1);
  }
  return n1 / m1;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  sort(a, a + n);
  int t = 0;
  int t1 = 0;
  b[0] = b[1] = b[2] = 1;
  for (int i = 1; i < n; i++) {
    if (a[i] == a[t1])
      b[t]++;
    else {
      t++;
      t1 = i;
    }
    if (t >= 3) break;
  }
  long long ans;
  if (b[0] >= 3)
    ans = cal(b[0], 3);
  else {
    if (b[1] + b[0] >= 3) {
      ans = cal(b[1], 3 - b[0]);
    } else {
      ans = cal(b[2], 3 - b[0] - b[1]);
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 maxn = 100000 + 10;
int a[maxn], b[3];
long long cal(int n, int m) {
  long long n1 = 1;
  long long m1 = 1;
  for (int i = n; i > n - m; i--) {
    n1 *= i;
    m1 *= (n - i + 1);
  }
  return n1 / m1;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  sort(a, a + n);
  int t = 0;
  int t1 = 0;
  b[0] = b[1] = b[2] = 1;
  for (int i = 1; i < n; i++) {
    if (a[i] == a[t1])
      b[t]++;
    else {
      t++;
      t1 = i;
    }
    if (t >= 3) break;
  }
  long long ans;
  if (b[0] >= 3)
    ans = cal(b[0], 3);
  else {
    if (b[1] + b[0] >= 3) {
      ans = cal(b[1], 3 - b[0]);
    } else {
      ans = cal(b[2], 3 - b[0] - b[1]);
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

```