#include <bits/stdc++.h>
const int MAXN = (int)1e5 + 5;
const int MODN = (int)1e9 + 7;
int inf = 0x3f3f3f3f;
using namespace std;
int a[MAXN];
long long C(int n, int m) {
  long long sum = 1;
  for (int i = 1; i <= m; i++) {
    sum *= (long long)n;
    sum /= (long long)i;
    n--;
  }
  return sum;
}
int main() {
  int n;
  int s1, s2, s3;
  int n1, n2, n3;
  n1 = n2 = n3 = 0;
  s1 = s2 = s3 = -1;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    if (s1 == -1) {
      s1 = a[i];
      n1 = 1;
    } else if (s1 == a[i]) {
      n1++;
    } else if (s2 == -1) {
      s2 = a[i];
      n2 = 1;
    } else if (s2 == a[i]) {
      n2++;
    } else if (s3 == -1) {
      s3 = a[i];
      n3 = 1;
    } else if (s3 == a[i]) {
      n3++;
    }
  }
  long long ans;
  if (n1 >= 3) {
    ans = C(n1, 3);
  } else if (n1 + n2 >= 3) {
    ans = C(n2, 3 - n1);
  } else if (n1 + n2 + n3 >= 3) {
    ans = C(n3, 3 - n1 - n2);
  }
  printf("%I64d\n", ans);
}

### 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>
const int MAXN = (int)1e5 + 5;
const int MODN = (int)1e9 + 7;
int inf = 0x3f3f3f3f;
using namespace std;
int a[MAXN];
long long C(int n, int m) {
  long long sum = 1;
  for (int i = 1; i <= m; i++) {
    sum *= (long long)n;
    sum /= (long long)i;
    n--;
  }
  return sum;
}
int main() {
  int n;
  int s1, s2, s3;
  int n1, n2, n3;
  n1 = n2 = n3 = 0;
  s1 = s2 = s3 = -1;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    if (s1 == -1) {
      s1 = a[i];
      n1 = 1;
    } else if (s1 == a[i]) {
      n1++;
    } else if (s2 == -1) {
      s2 = a[i];
      n2 = 1;
    } else if (s2 == a[i]) {
      n2++;
    } else if (s3 == -1) {
      s3 = a[i];
      n3 = 1;
    } else if (s3 == a[i]) {
      n3++;
    }
  }
  long long ans;
  if (n1 >= 3) {
    ans = C(n1, 3);
  } else if (n1 + n2 >= 3) {
    ans = C(n2, 3 - n1);
  } else if (n1 + n2 + n3 >= 3) {
    ans = C(n3, 3 - n1 - n2);
  }
  printf("%I64d\n", ans);
}

```