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>
#define ll long long int
using namespace std;
bool ispal(string a)
{
    long long int n=a.length();
    if(n==1)
        return true;
    else{
    int i=0,j;
    j=n-1;
    while(1)
    {
        if(a[i]!=a[j])
        {
            cout<<i<<" "<<j;
           return false;
        }
        i++,j--;
        if(i==j)
         {
        //     cout<<i<<" ";
             break;
        }
    }
    return true;
}}

int main(){
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ll n;
  cin>>n;
  ll a[n];
  for(int i=0;i<n;i++)
  {
      cin>>a[i];
  }
  sort(a,a+n);
ll  c=0;
  for(int i=0;i<n-2;i++)
  {
      int as=a[i];
      for(int j=i+1;j<n-1;j++)
      {
          int bs=a[j];
          for(int k=j+1;k<n;k++)
          {

            if(a[k]<as+bs)
            {
                if(a[k]+as>bs)
                {
                    if(a[k]+bs>as)
                        {
                            if(as!=bs && bs!=a[k] && a[k]!=as)
                            c++;
                        }
                }
            }
          }

      }

  }
  cout<<c;
  }