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##_len = (n); i < i##_len; ++i)
#define reps(i, n) for (int i = 1, i##_len = (n); i <= i##_len; ++i)
#define rrep(i, n) for (int i = ((int)(n)-1); i >= 0; --i)
#define rreps(i, n) for (int i = ((int)(n)); i > 0; --i)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)
#define repc2(i, s, n) for (int i = (s); i <= (int)(n); i++)
#define inf 2e9
#define linf 9000000000000000000ll
#define all(v) v.begin(), v.end()
using namespace std;
typedef long long ll;

ll n;
vector<ll> v;

int main() {
    int ans = 0;
    cin >> n;
    v.resize(n);
    rep(i, n) cin >> v[i];
    rep(i, n) rep2(j, i + 1, n) rep2(k, j + 1, n) {
        ll sum = v[i] + v[j] + v[k];
        ll mx = max({v[i], v[j], v[k]});
        if ((sum - mx > mx) && v[i] != v[j] && v[j] != v[k] && v[i] != v[k]) ans++;
    }
    cout << ans << endl;
    return 0;
}