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)


//auto 関数名 = [(&:ラムダ外の引数をとる時))](引数の型1 引数名1, 引数の型2, 引数名2, ...) { 関数の処理 }; //inside main() define function. take care of };
// for (int tmp = 0; tmp < (1 << ビット数); tmp++) {
//   bitset<ビット数> s(tmp);
//   // (ビット列sに対する処理)
// }
// sort(配列変数.begin(), 配列変数.end());
// do {
//   // 順列に対する処理
// } while (next_permutation(配列変数.begin(), 配列変数.end()));
//const double PI = acos(-1); M_PI
// cout << fixed << setprecision(10);
#include <bits/stdc++.h>
#include <regex>
using namespace std;
#define _GLIBCXX_DEBUG 
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define all(v) v.begin(), v.end() //sort(all(vec)); ,reverse(all(vec));etc
int main(){
  int n;
  cin>>n;
  vector<int>vec(n);
  int l;
  rep(i,n){
    cin>>vec.at(i);
  }
  int ans=0;
  sort(all(vec));
  rep(i,n-2){
    rep(j,n-i-1){
      rep(k,n-i-j-2){
        //cout<<vec.at(i)<<" "<<vec.at(j+i+1)<<" "<<vec.at(k+i+j+2)<<endl;
        if(vec.at(i)==vec.at(j+i+1)||vec.at(j+i+1)==vec.at(k+i+j+2)){
          continue;
          
        }
        if(vec.at(i)+vec.at(j+i+1)<=vec.at(k+i+j+2)){
         
          continue;
        }
        ans+=1;
      }
    }
  }
  cout<<ans<<endl;
}