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;
using ull = unsigned long long;
using ld = long double;

int main(){

    int N;
    cin >> N;
    vector<ull> L(N);
    for(int i=0; i<N; i++) cin >> L.at(i);

    ull a = 0, b = 0, c = 0, cnt = 0;
    for(int i=0; i<N-2; i++){
        for(int j=i+1; j<N-1; j++){
            for(int k=j+1; k<N; k++){
                a = L.at(i);
                b = L.at(j);
                c = L.at(k);

                if ((a+b > c) && (b+c > a) && (c+a > b) && (L.at(i) != L.at(j)) && (L.at(j) != L.at(k)) && (L.at(k) != L.at(i))){
//                    cerr << i+1 << " " << j+1 << " " << k+1 << endl;
//                    cerr << L.at(i) << " " << L.at(j) << " " << L.at(k) << endl;
//                    cerr << a << " " << b << " " << c << endl;
                    cnt++;
                }
            }
        }
    }
    cout << cnt << endl;
    return 0;
}