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)

//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
#define all(v) v.begin(), v.end()
using in = int64_t;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define REP(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define Yes cout<<"Yes"<<endl
#define No cout<<"No"<<endl
#define yes cout<<"yes"<<endl
#define no cout<<"no"<<endl
const double PI=3.14159265358979323846;
const in MOD = 1000000007;
const in INF=1e18+7;
const int inf=1e9+7;
using P=pair<int,int>;
vector<int> dx={0,1,-1,0};
vector<int> dy={1,0,0,-1};


int main(){
ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
cout << fixed << setprecision(10);
    int n;
    cin>>n;
    vector<int> a(n);
        rep(i,n) cin>>a.at(i);
    int cnt=0;
    rep(i,n-2){
        REP(j,i+1,n-1){
            if(a[i]==a[j]) continue;
            REP(k,j+1,n){
                if(a[i]==a[k] ||a[j]==a[k]) continue;
                if(abs(a[i]-a[j])>=a[k] || a[i]+a[j]<=a[k]) continue;
                cnt ++;
            }
        }
    }
    cout<<cnt<<endl;
}/*
./problem.exe
*/