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 sws ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define ll long long
#define vi vector<int>
#define pb push_back
#define pii pair<int, int>
#define mp make_pair
#define ff first
#define ss second
#define all(x) x.begin(), x.end()
#define sz(x) ((int) x.size())
#define fo(i, n) for( int i = 0; i < n; i++ )

const int INF = 1e9;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1e9 + 7;


int main() {sws;

    int n;
    ll l[105];
    cin >> n;
    fo(i, n) cin >> l[i];

    int ans = 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[k] == l[i] or l[k] == l[j] ) continue;
                int a = l[i], b = l[j], c = l[k];

                if ( abs(b - c) < a and a < b + c and abs(a - c) < b and b < a + c and abs(a - b) < c and c < a + b ) ans++;
            }
        }
    }

    cout << ans << endl;

    return 0;
}