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 
#define MM(arr,x) memset(arr,x,sizeof(arr))
#define INF 0x3f3f3f3f
#define Maxn 1e5+10;
using namespace std;
int a[205];
int main()
{
	ios::sync_with_stdio(0);
	int n;
	cin>>n;
	for(int i=0;i<n;i++)
		cin>>a[i];
	int cnt = 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(a[i]!=a[j]&&a[i]!=a[k]&&a[j]!=a[k])
				{
					int min1 = min(a[i],min(a[j],a[k]));
					int min2 = a[i]+a[j]+a[k]-min1-max(a[i],max(a[j],a[k]));
					int max1 = max(a[i],max(a[j],a[k]));
					if(min1+min2>max1&&min2-min1<max1) cnt++;
				}
			}
		}
	}
	cout<<cnt;
	return 0;
}