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<bits/stdc++.h>
using namespace std;
int fact(int n) 
{ 
    int res = 1; 
    for (int i = 2; i <= n; i++) 
        res = res * i; 
    return res; 
} 

int nCr(int n, int r) 
{ 
    return fact(n) / (fact(r) * fact(n - r)); 
} 

int   main(){
   int n;
   cin>>n;
   int temp;
   vector<int> a;
   for(int i=0;i<n;++i){
       cin>>temp;
       a.push_back(temp);
   }
   sort( a.begin(), a.end() );
// a.erase( unique( a.begin(), a.end() ), a.end() );
//  n = a.size();
 int ans= 0 ;
  for(int i=0;i<n-2;++i){
      for(int j=i+1;j<n-1;++j){
          for(int k=j+1;k<n;++k){
              if(a[i]+a[j]>a[k]&&a[i]!=a[j]&&a[j]!=a[k]&&a[k]!=a[i]){
                  ans++;
              }
          }
      }
  }
   
   cout<<ans;
  
    return 0;

}