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>
#include <vector>
#include <iostream>
#include<algorithm>
#include<string>
#include <map>
#include <queue>
#include <stack>
#include<set>

#define DIV 1000000007
#define TE 2e5
using namespace std;
using ll = long long;
using ldb = long double;

int main() {
	int N; cin >> N;
	vector<ll> L(N);
	for (int i = 0; i < N; i++) {
		cin >> L[i];
	}

	int cnt = 0;
	for (int i = 0; i < N; i++) {
		for (int j = i + 1; j < N; j++) {
			if (L[i] == L[j])continue;
			for (int k = j + 1; k < N; k++) {
				if (L[i] == L[k] || L[j] == L[k])continue;
				if (abs(L[i] - L[j]) < L[k] && L[k] < abs(L[i] + L[j]))cnt++;
			}
		}
	}
	cout << cnt << endl;

}