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 <iostream>
#include <vector>
#include <string>
#include <list>
#include <queue>
#include <algorithm>
#define rep(i, n) for(i = 0; i < (n); i++)
#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
#define MOD 1000000007
#define PI 3.14159265358979323846
#define INF 1 << 30

using namespace std;
typedef long long ll;
typedef pair<int, int> pp;

int Solve(vector<int> & d, int s) {
	int i, j, ans = 0;
	rep(i, s) {
		if (d[i] == d[s])
			continue;
		for (j = i + 1; j < s; j++) {
			if (d[i] == d[j] || d[j] == d[s])
				continue;
			if(d[i] + d[j] > d[s])
				ans++;
		}
	}
	return ans;
}

int main(void) {
	int num, i;
	ll ans = 0;
	cin >> num;
	vector<int> d(num);
	rep(i, num)
		cin >> d[i];
	sort(d.begin(), d.end());
	for (i = 2; i < d.size(); i++) {
		ans += Solve(d, i);
	}
	cout << ans << "\n";
	return 0;
}