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;

#define all(x) (x).begin(),(x).end()

using ll = long long;
using P = pair<int,int>;
using mp =  map<string,int>;

const int MOD = 1e9 + 7;
const int INF = 1001001001;


int main(void){

    int n;
    cin >> n;

    vector<int> a(n);
    for(int i = 0; i < n; ++i){
        cin >> a[i];
    }

    int ans = 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[k] && a[i] + a[k] > a[j] && a[k] + a[j] > a[i] && a[i] != a[j] && a[j] != a[k] && a[i] != a[k]){
                    ans++;
                }
            }
        }
    }

    cout << ans << "\n";

    return 0;
}