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;
 
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef pair<ll,ll> P;
typedef string str;
typedef vector<P> vp;
typedef vector<string> vs;
typedef vector<bool> vb;

const ll mod=1e9;
const ll inf=1e16;

#define rep(i,m,n) for(ll i=m;i<n;i++)
#define repr(i,m,n) for(ll i=m-1;i>=n;i--)
#define fi first
#define se second
#define chmax(x,y) x=max(x,y)
#define chmin(x,y) x=min(x,y)
#define eb(x) emplace_back(x)
#define pb(x) pop_back(x)
#define all(x) x.begin(),x.end()
#define allr(x) x.rbegin(),x.rend()
#define sum(x) accumulate(all(x),0)
#define pc(x) __builtin_popcount(x)
#define gll greater<ll>()

void solve(){
 ll n;
  cin >> n;
 vll l(n);
  rep(i,0,n) cin >> l[i];
  sort(all(l));
 
  ll ans=0;
  rep(i,0,n){
   rep(j,i+1,n){
     if(l[i]==l[j]) continue;
     rep(k,j+1,n){
      if(l[j]==l[k]) continue;
      if(abs(l[j]-l[i])<l[k]&&l[k]<l[i]+l[j]) ++ans;
     }
   }
  }
  cout << ans << endl;
}
int main(){
   cin.tie(nullptr);
  ios::sync_with_stdio(false);
  solve();
}