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;

typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
const double pi = 3.14159265358979323846;

#define all(s) s.begin(),s.end()
#define speed ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)

void rw()
{
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif // ONLINE_JUDGE
}

int main()
{
	speed;
	//rw();

	ll n, x;
	cin >> n;

	vector<ll> r;
	for (int i = 0; i < n; i++)
		cin >> x, r.push_back(x);

	sort(all(r));
	x = 0;
	if (r.size() > 2)
	{
		n = r.size();
		for (int i = 0; i < n; i++)
		{
			for (int j = i + 1; j < n; j++)
			{
				for (int k = j + 1; k < n; k++)
					if (r[i] + r[j] > r[k] && r[i] != r[j] && r[j] != r[k] && r[k] != r[i])
						x++;
			}
		}
	}
	else
		x = 0;

	cout << x;
	return 0;
}