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;
using ll = long long;
int gcd(int a,int b)
{
	if(b==0)
	return a;
	
	return gcd(b,a%b);
}
int power(int x,int y,int p)
{
	int res=1;
	x=x%p;
	while(y>0)
	{
		if(y&1)
		res=(res*x)%p;
		
		y=y>>1;
		x=(x*x)%p;
	}
	return res;
}
struct Matrix{ 
	int a[2][2] = {{0,0},{0,0}};
	Matrix operator *(const Matrix& other){
		Matrix product;
		for(int i=0;i<2;i++){
			for(int j=0;j<2;j++){
				for(int k=0;k<2;k++){
						product.a[i][k] =product.a[i][k] + (ll)a[i][j]*other.a[j][k];
				}
			}
		}
		return product;
	}
};
int modInverse(int n,int p)
{
	return power(n,p-2,p);
}
int main() {
	// your code goes here
	 ios_base::sync_with_stdio(0);
	 cin.tie(0);
	 int n;
	 cin>>n;
	 vector<int> v(n);
	 for(int i=0;i<n;i++)
	 {
		 cin>>v[i];
	 }
	 ll ans = 0;
	 for(int i=0;i+2<n;i++)
	 {
		 for(int j=i+1;j+1<n;j++)
		 {
			 for(int k=j+1;k<n;k++)
			 {
				 if(v[i]!=v[j] && v[j]!=v[k] && v[i]!=v[k])
				 {
					 if(v[i]+v[j]>v[k] && v[i]+v[k]>v[j] && v[j]+v[k]>v[i])
					 ans++;
				 }
			 }
		 }
	 }
	 cout<<ans<<'\n';
	 return 0;
}