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 main()
{
	long long int n;
	cin>>n;
	vector<long long int> v;
	for(int i=0;i<n;i++)
	{
		long long int x;
		cin>>x;
		v.push_back(x);
	}
	long long int ans=0;
	for(int i=0;i<n;i++)
	{
		for(int j=i+1;j<n;j++)
		{
			for(int k=j+1;k<n;k++)
			{
				if(v[i]!=v[j]&&v[i]!=v[k]&&v[j]!=v[k])
				{
					int m;
					m=max(v[i],max(v[j],v[k]));
					if(m==v[i])
					{
						if((v[j]+v[k])>v[i])
						{
							ans++;
						}
					}
					if(m==v[j])
					{
						if((v[i]+v[k])>v[j])
						{
							ans++;
						}
					}
					if(m==v[k])
					{
						if((v[j]+v[i])>v[k])
						{
							ans++;
						}
					}
				}
			}
		}
	}
	cout<<ans<<endl;
}