Problem:
Score : 200 points

Problem Statement

We have sticks numbered 1, \cdots, N. The length of Stick i (1 \leq i \leq N) is L_i.

In how many ways can we choose three of the sticks with different lengths that can form a triangle?

That is, find the number of triples of integers (i, j, k) (1 \leq i < j < k \leq N) that satisfy both of the following conditions:

L_i, L_j, and L_k are all different.

There exists a triangle whose sides have lengths L_i, L_j, and L_k.

Constraints

1 \leq N \leq 100

1 \leq L_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
L_1 L_2 \cdots L_N

Output

Print the number of ways to choose three of the sticks with different lengths that can form a triangle.

Sample Input 1

5
4 4 9 7 5

Sample Output 1

5

The following five triples (i, j, k) satisfy the conditions: (1, 3, 4), (1, 4, 5), (2, 3, 4), (2, 4, 5), and (3, 4, 5).

Sample Input 2

6
4 5 4 3 3 5

Sample Output 2

8

We have two sticks for each of the lengths 3, 4, and 5. To satisfy the first condition, we have to choose one from each length.

There is a triangle whose sides have lengths 3, 4, and 5, so we have 2 ^ 3 = 8 triples (i, j, k) that satisfy the conditions.

Sample Input 3

10
9 4 6 1 9 6 10 6 6 8

Sample Output 3

39

Sample Input 4

2
1 1

Sample Output 4

0

No triple (i, j, k) satisfies 1 \leq i < j < k \leq N, so we should print 0.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
using namespace std;
int main(){
  int N;
  int arr[100];
  cin >> N;
  for(int i=0; i < N; i++){
  	cin >> arr[i]; 
  }
  
  int answer = 0;
  
  for(int i=0; i < N-2; i++)
    for(int k=i+1; k < N-1; k++)
      for(int j=k+1; j < N; j++)
        if(arr[i]!=arr[k] && arr[k]!=arr[j] && arr[i]!=arr[j]){
          int tmp = arr[i]>arr[k] ? arr[i] : arr[k];
          tmp = tmp > arr[j] ? tmp : arr[j];
          int tmp2 = arr[i]+arr[k]+arr[j]-tmp-tmp;
          if(tmp2 > 0) answer++;
        }
  cout << answer;
}