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 _ ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define int long long
const int inf = 0x3f3f3f3f3f3f3f3LL;
const int mod = (int)1e9 + 7;
using namespace std;
#if LOCAL
#include "../../tr.h"
#else
#define tr(...)
#define endl '\n'
#endif
template<class T, class T2> void smin(T& a, T2 val) {if (a > val) a = val;}template<class T, class T2> void smax(T& a, T2 val) {if (a < val) a = val;}
const int N = 5*(int)1e5 + 10;

int n, a[N];

int32_t main(){_
   
   cin>>n;

   for (int i = 0; i < n; ++i)
   {
      cin>>a[i];
   }

   int ans = 0;
   for (int i = 0; i < n; ++i)
   {
      for (int j = i+1; j < n; ++j)
      {
         for (int k = j+1; k < n; ++k)
         {
            if (a[i] == a[j] or a[i] == a[k] or a[j] == a[k]){
               continue;
            }
            if (a[j] + a[k] > a[i] and a[k] + a[i] > a[j] and a[j] + a[i] > a[k]){
               tr(i+1, j + 1, k+1);
               ++ans;
            }
         }
      }
   }
   
   cout<<ans;

   return 0;
}