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 ll long long
#define ull unsigned long long
#define pii pair<int,int>
#define pp pair<pair<ll, ll>,pair<ll, ll>>
#define pll pair<ll,ll>
#define ppll pair<ll,pll>
#define pdd pair<double,double>
#define vii vector<int>
#define vll vector<ll>
#define mat vector<vector<ll>>
#define lb lower_bound
#define ub upper_bound
#define pb push_back
#define eb emplace_back
#define fi first
#define sc second
#define rep(i,n) for(ll i=0;i<n;i++)
#define rep2(i,a,b) for(ll i=a;i<b;i++)
#define repr(i,n) for(ll i=n-1;i>=0;i--)
#define all(x) x.begin(),x.end()
#define sz(x) (ll) (x).size()
#define pq priority_queue<ll>
#define pqg priority_queue<ll,vector<ll>,greater<ll>>
#define LB(v,x) (lower_bound(v.begin(),v.end(),x)-v.begin())
#define UB(v,x) (upper_bound(v.begin(),v.end(),x)-v.begin())
#define ERASE(v) sort(v.begin(),v.end());v.erase(unique(v.begin(),v.end()),v.end())
#define int ll
// #define ll int
using namespace std;
const ll INF = 1 << 29;
const ll LLINF = (1LL << 60LL);
const ll MOD = 1000000007;
const int MAX_V = 500;
// const ll mod = 998244353;
const ll MAX = 2100000;
const double pi = acos(-1);
const double eps = 1e-10;
ll dx[8] ={1,0,-1,0,1,-1,1,-1};
ll dy[8] ={0,1,0,-1,1,-1,-1,1};
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }







void solve() {
    ll n;
    ll a[100010];
    cin>>n;
    rep(i,n) cin>>a[i];
    ll ans=0;
    rep(i,n){
        rep2(j,i+1,n){
            rep2(k,j+1,n){
                ll sum=a[i]+a[j]+a[k];
                if(a[i]!=a[j] && a[j]!=a[k] && a[i]!=a[k]){
                    ll s=max(a[i],max(a[j],a[k]));
                    if(sum-s > s) ans++;
                }
            }
        }
    }
    cout<<ans<<endl;

}






signed main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);


    ll t=1;
    rep(i,t){
        solve(); 
    }


    return 0;
}