#include <bits/stdc++.h>
using namespace std;
int factorial(int n) { return (n == 1 || n == 0) ? 1 : factorial(n - 1) * n; }
int main() {
  int N = 0;
  cin >> N;
  int numlist[100000];
  for (int i = 0; i < N; i++) {
    cin >> numlist[i];
  }
  sort(numlist, numlist + N);
  int x = 1;
  if (numlist[0] == numlist[1]) {
    x++;
  }
  if (numlist[1] == numlist[2]) {
    x++;
  }
  int a1 = numlist[0];
  int a2 = numlist[1];
  int a3 = numlist[2];
  long long count1 = 0;
  long long count2 = 0;
  long long count3 = 0;
  for (int i = 0; i < N; i++) {
    if (numlist[i] == a1) {
      count1++;
    }
  }
  for (int i = 1; i < N; i++) {
    if (numlist[i] == a2) {
      count2++;
    }
  }
  for (int i = 2; i < N; i++) {
    if (numlist[i] == a3) {
      count3++;
    }
  }
  cout << count1 * count2 * count3 / factorial(x);
}

### 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 factorial(int n) { return (n == 1 || n == 0) ? 1 : factorial(n - 1) * n; }
int main() {
  int N = 0;
  cin >> N;
  int numlist[100000];
  for (int i = 0; i < N; i++) {
    cin >> numlist[i];
  }
  sort(numlist, numlist + N);
  int x = 1;
  if (numlist[0] == numlist[1]) {
    x++;
  }
  if (numlist[1] == numlist[2]) {
    x++;
  }
  int a1 = numlist[0];
  int a2 = numlist[1];
  int a3 = numlist[2];
  long long count1 = 0;
  long long count2 = 0;
  long long count3 = 0;
  for (int i = 0; i < N; i++) {
    if (numlist[i] == a1) {
      count1++;
    }
  }
  for (int i = 1; i < N; i++) {
    if (numlist[i] == a2) {
      count2++;
    }
  }
  for (int i = 2; i < N; i++) {
    if (numlist[i] == a3) {
      count3++;
    }
  }
  cout << count1 * count2 * count3 / factorial(x);
}

```