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 <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <iostream>
#include <iostream>
#include <fstream>
#include <numeric>
#include <cstring>
#include <cassert>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <cmath>
#include <set>
#include <map>
#include <functional>
#include <bitset>
#include <iomanip>
#include <stack>
#include <list>
using namespace std;
#define ll           long long
#define all(v)       (v).begin() , (v).end()
#define vi           vector <ll>
#define vii          vector <pair<ll,ll>>
#define ii           pair<ll,ll>
#define sz(v)	     ((int)((v).size()))
#define lp(i,a,b)    for(int i=a;i<b;i++)
#define pb           push_back
#define pf           push_front
#define F            first
#define S            second
#define endl         "\n"
#define IOS          ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
const double PI = 3.14159265358979323846;
const ll inf=2e9,MOD=998244353,MAX=1e5+1;

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    IOS;
    int n;
    cin>>n;
    int a[n];
    lp(i,0,n)
        cin>>a[i];
    int cnt=0;
    lp(i,0,n){
        lp(j,i+1,n){
            lp(k,j+1,n){
                if(a[i]!=a[j]&&a[i]!=a[k]&&a[j]!=a[k]&&a[i]+a[j]>a[k]&&a[i]+a[k]>a[j]&&a[j]+a[k]>a[i])
                    cnt++;
            }
        }
    }
    cout<<cnt;
}