#include <bits/stdc++.h>
using namespace std;
int main() {
  int nb;
  scanf("%d", &nb);
  int arr[nb];
  for (int i = 0; i < nb; i++) scanf("%d", arr + i);
  sort(arr, arr + nb);
  int occ[3] = {0};
  for (int i = 0; i < nb; i++) {
    for (int j = 0; j < 3; j++)
      if (arr[i] == arr[j]) occ[j]++;
  }
  if ((occ[0] == 1 && occ[1] == 1) || (occ[0] == 2)) {
    printf("%d", occ[2]);
    return 0;
  }
  if (occ[0] == 1 && occ[1] >= 2) {
    long long res = (occ[1] * (long long)(occ[1] - 1)) / 2;
    printf("%lld", res);
    return 0;
  }
  long long res = (occ[0] * (long long)(occ[0] - 1) * (occ[0] - 2)) / 6;
  printf("%lld", res);
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
int main() {
  int nb;
  scanf("%d", &nb);
  int arr[nb];
  for (int i = 0; i < nb; i++) scanf("%d", arr + i);
  sort(arr, arr + nb);
  int occ[3] = {0};
  for (int i = 0; i < nb; i++) {
    for (int j = 0; j < 3; j++)
      if (arr[i] == arr[j]) occ[j]++;
  }
  if ((occ[0] == 1 && occ[1] == 1) || (occ[0] == 2)) {
    printf("%d", occ[2]);
    return 0;
  }
  if (occ[0] == 1 && occ[1] >= 2) {
    long long res = (occ[1] * (long long)(occ[1] - 1)) / 2;
    printf("%lld", res);
    return 0;
  }
  long long res = (occ[0] * (long long)(occ[0] - 1) * (occ[0] - 2)) / 6;
  printf("%lld", res);
  return 0;
}

```