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;
#define rep(i, n) for(ll i = 0, i##_len = (n); i < i##_len; ++i)
#define rep2(i, x, n) for(ll i = x, i##_len = (n); i < i##_len; ++i)
#define all(n) begin(n), end(n)
using ll = long long;
using P = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vs = vector<string>;
using vc = vector<char>;
using vb = vector<bool>;
using vd = vector<double>;
vi dir = {-1, 0, 1, 0, -1, -1, 1, 1, -1};

int main() {
  ll n;
  cin >> n;
  vl a(n);
  rep(i, n) cin >> a[i];
  ll ans = 0;
  rep(i, n - 2) rep2(j, i + 1, n - 1) rep2(k, j + 1, n) {
    if(a[i] == a[j] || a[j] == a[k] || a[k] == a[i]) continue;
    vl x(3);
    x[0] = a[i];
    x[1] = a[j];
    x[2] = a[k];
    sort(all(x));
    if(x[0] + x[1] <= x[2]) continue;
    ans++;
  }
  cout << ans << endl;
}