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)

#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("fast-math")
#pragma GCC optimize("trapv")
#pragma GCC target("sse4")

#include <bits/stdc++.h>
using namespace std;

#define deb(x) cout << #x << " is " << x << "\n"
#define int long long
#define mod 1000000007
#define PI acos(-1)

void solve() {
	int n;
	cin >> n;
	vector<int> arr(n);
	for(auto &ele : arr)
		cin >> ele;
	int ans = 0;
	for(int i = 0; i < n; i++) {
		for(int j = i + 1; j < n; j++) {
			for(int k = j + 1; k < n; k++) {
				vector<int> temp = {arr[i], arr[j], arr[k]};
				sort(temp.begin(), temp.end());
				if(arr[i] != arr[j] && arr[j] != arr[k] && arr[i] != arr[k] && temp[0] + temp[1] > temp[2]) {
					ans++;
				} 
			}
		}
	}
	cout << ans;
}

int32_t main() {
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
// #ifndef ONLINE_JUDGE
// 	freopen("input.txt", "r", stdin);
// 	freopen("output.txt", "w", stdout);
// #endif
	int tc = 1;
	// cin >> tc;
	while(tc--) {
	    solve();
	}
	return 0;
}