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;

// #include <ext/pb_ds/assoc_container.hpp> 
// #include <ext/pb_ds/tree_policy.hpp> 
// using namespace __gnu_pbds; 
  
// #define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update> 

#define ll long long
#define fill fill_n
#define pf push_front
#define pb push_back
#define FOR(i,l,r) for(int i=l;i<r;i++)
#define ROF(i,r,l) for(int i=r;i>=l;i--)
#define comp(v) (v).begin(),(v).end()
#define ff first
#define ss second
#define infile freopen("input.txt", "r", stdin);freopen("output.txt", "w", stdout);
typedef vector<int> vi;
typedef vector<ll> vli;
typedef vector<pair<int,int>> vpi;

const int M = 1e9+7;
const int N = 2e5;
ll po(int,int);


void solve(){
   int n;cin>>n;
   map<ll,ll> m;
   ll l;
   FOR(i,0,n){
       cin>>l;
       m[l]++;
   }
   ll ans=0;
   for(auto i:m){
       for(auto j:m){
           if(j.ff>i.ff){
               for(auto k:m){
                   if(k.ff>j.ff && k.ff>i.ff && i.ff+j.ff > k.ff){
                       ans += (i.ss*j.ss*k.ss);
                   }
               }
           }
       }
   }
   cout<<ans<<"\n";
    
}





int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
    int t=1;
    //cin>>t;
    FOR(i,1,t+1){
        //cout<<"Case #"<<i<<": "; 
        solve();
    }
    
	return 0;
	
	
}


ll po(int a,int b){
    if(b==0) return 1;
    ll ans=1;
    if(b%2==0){
        ans = po(a,b/2)%M;
        ans=(ans%M * ans%M)%M;
    }
    else{
        ans = po(a,(b-1)/2)%M;
        ans = (ans%M * ans%M * a)%M;
    }
    return ans%M;
}