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;
#include<bits/stdc++.h>
using namespace std;
typedef     long long    ll;
typedef     vector<int> vi;
typedef     vector<long long> vl;
typedef     pair<int, int>pi;
typedef     pair<long long, long long>pl;
#define F   first
#define S   second
#define pb  push_back
#define     all(x)      x.begin() , x.end()
#define mp  make_pair
#define       FOR(i,a,b) for(i=a;i<=b;i++)
#define     mem(a)      memset(a , 0 ,sizeof a)
#define     memn(a)     memset(a , -1 ,sizeof a)
int main()
{

  int n, c = 0, j, x, y,k,i;
  cin >> n;
  int a[n + 3];
  int b[10];
  for (i = 0; i < n; i++)
  {
    cin >> a[i];
  }
  for (i = 0; i < n - 2; i++)
  {
    for (j = i + 1; j < n - 1; j++)
    {
      for (k = j + 1; k < n; k++)
      {
        if (a[i] != a[j] && a[j] != a[k] && a[i] != a[k])
        {
          b[0] = a[i]; b[1] = a[j]; b[2] = a[k];
          sort(b, b + 3);
          if (b[0] + b[1] > b[2])
          {
            c++;
          }
        }
      }
    }
  }

  cout<<c<<endl;
}