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>
 
#define f(i,a,b) for( ll i = a; i < b ; i++ )
#define af(i,a,b) for( ll i = a; i >= b ; i--)
#define rep(i,a,b,k) for(ll i = a; i < b ; i+= k )
#define arep(i,a,b,k) for( ll i = a; i >= b ; i-= k)
#define ones(x) (ll) __builtin_popcount(x)
#define fs first
#define sc second
#define pb push_back
#define po pop_back
#define mp make_pair  
#define sz(a) (ll) a.size()
#define all(a) a.begin(), a.end()
#define sor(a) sort( a.begin(), a.end() )
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL)
#define ller ios::sync_with_stdio(false);cin.tsie(nullptr);cout.tie(nullptr)
#define watch(x) cout << (#x) << " is #" << (x) <<"#" <<endl
#define PI 3.1415926535
using namespace std;
typedef long long ll;
typedef long double ld; 
typedef pair<ll,ll> ii ;
typedef vector<ll>  vi ;
typedef vector<ii> vii ;
const ll mod = 1e9+7;
const ll MAX = 1e2+5;
const ll inf = 1e18+7;

ll a[MAX],cont[MAX];

int main() {
  fastio;
  ll n;
  cin>>n;
  f(i,0,n){
      cin>>a[i];
  }
  sort(a,a+n);
  ll ans = 0;
  f(i,0,n){
    f(j,i+1,n){
      f(k,j+1,n){
        ll x = a[i];
        ll y = a[j];
        ll z = a[k];
        if(x!=y&&y!=z&& x+y>z){ 
          ans++;
        //  cout<<"TRIANGULO "<<i+1<<" "<<j+1<<" "<<k+1<<"\n";
        }
      }
    }
  }
  cout<<ans<<"\n";
 return 0;
}