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 pb              push_back
#define endl            "\n"
#define sz(x)          (int) (x.size())
#define all(v)          v.begin(), v.end()
#define isOdd(n)       (n&1)
#define pow(n, m)      (ll)powl(n, m)
#define clr(x, val)     memset(x, val, sizeof(x))

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef vector<int> vi;

bool can(int a, int b, int c){
    vector<int> v = {a, b, c};
    sort(all(v));
    if(v[0] == v[2] || v[0] == v[1] || v[1] == v[2]) return 0;
    return v[0] + v[1] > v[2];
}
int main()
{
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

    int n; cin >> n;
    int a[n];
    ll c = 0;
    for(auto &i: a) cin >> i;
    for(int i = 0; i < n; i++){
        for(int j = i + 1; j < n; j++){
            for(int k = j + 1; k < n; k++){
                if(can(a[i], a[j], a[k])){
                    c++;
                }
            }
        }
    }
    cout << c;
    return 0;
}