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 rep(i, n) for(int i = 0; i < (int)(n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;

const ll MOD = 1e9+7;

int main() {
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  
  int n;
  cin >> n;
  vector<int> l(n);
  rep(i, n) cin >> l[i];
  int cnt = 0;
  rep(i, n) {
    int a, b, c;
    rep(j, n) {
      if(j <= i) continue;
      rep(k, n) {
        if(k <= j) continue;  
        if(l[i] != l[j] && l[j] != l[k] && l[k] != l[i]) {
          a = l[i];
          b = l[j];
          c = l[k];
          if(a+b > c && b+c > a && a+c > b) ++cnt;
        }
      }
    }
  }
  
  cout << cnt << endl;
  return 0;
}