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>
#include <vector>
#define rep(i, n) for(int i = 0; i < n; ++i)
#define ll long long
#define all(a) a.begin(), a.end() 
using namespace std;

int main() {
  int n;
  cin >> n;
  vector<int> ip(n);
  rep(i, n) cin >> ip[i];
  sort(ip.begin(), ip.end());
  vector<ll int> len;
  int last = 0;
  map<int, ll int> mp;
  for(int x : ip) {
    ++mp[x];
    if(last != x) len.push_back(x);
    last = x;
  }
  
  int m = len.size();
  // rep(i, m) cout << len[i] << " ";
  
  int ans = 0;
  for(int i = 0; i < m-2; ++i) {
    for(int j = i+1; j < m-1; ++j) {
      for(int k = j+1; k < m; ++k) {
        if(len[i] + len[j] > len[k]) {
          ans += mp[len[i]]*mp[len[j]]*mp[len[k]];
        }
      }
    }
  }
  
  cout << ans << endl;
}