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)

/* Author : tushar3105 */

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;

typedef pair<long long int,long long int> ii;					// Pair of integers
typedef vector<long long int> vi;                   				// Vector of integers
typedef vector<vector<long long int>> vvi;                 			// Vector of vi
typedef vector<pair<long long int, long long int>> vii;				// Vector of ii

#define pb push_back								// Push back to vectors
#define mp make_pair								// For pairs
#define ff first								// For pairs
#define ss second								// For pairs
#define all(c) (c).begin(),(c).end()						// Use in sorting
#define sz(a) ((long long int)(a).size())
#define lli long long int
#define ull unsigned long long int
#define ld long double
#define ref(i,x,y) for(long long int i=(x);i<=(y);++i)				// Loop to go forward
#define reb(i,x,y) for(long long int i=(x);i>=(y);--i)				// Loop to go backward
#define trf(c,it) for(auto it = (c).begin(); it != (c).end(); it++)		// Remember it is an iterator
#define trb(c,it) for(auto it = (c).end()-1; it != (c).begin()-1; it--)		// Remember it is an iterator
#define tc(t) long long int t; cin>>t;while(t--)				// Test cases
//fill (dist,dist+110000,pinf);							// For setting default

const long long int mod = 1e9+7;
const long long int pinf = 9223372036854775807;
const long long int ninf = -9223372036854775807;

/*-----------------------------Code begins----------------------------------*/

int main (void)
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	
	lli n, cnt=0, arr[110];
	
	cin>>n;
	
	ref(i,0,n-1)
		cin>>arr[i];
		
	ref(i,0,n-1)
	{
		ref(j,0,n-1)
		{
			ref(k,0,n-1)
			{
				if (arr[i]!=arr[j] && arr[j]!=arr[k] && arr[k]!=arr[i])
				{
					if (arr[i]+arr[j]>arr[k] && arr[j]+arr[k]>arr[i] && arr[k]+arr[i]>arr[j])
						cnt++;
				}
			}
		}
	}
	
	cout<<cnt/6<<endl;

	return 0;
}