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 <functional>
#include <vector>
using namespace std;

void recursive_comb(int* indexes, int s, int rest, std::function<void(int*)> f) {
	if (rest == 0) {
		f(indexes);
	}
	else {
		if (s < 0) return;
		recursive_comb(indexes, s - 1, rest, f);
		indexes[rest - 1] = s;
		recursive_comb(indexes, s - 1, rest - 1, f);
	}
}

void foreach_comb(int n, int k, std::function<void(int*)> f) {
	vector<int> indexes(k);
	recursive_comb(&(indexes[0]), n - 1, k, f);
}

int main()
{
	auto& in = cin;

    int N;
    in >> N;
	if (N < 3) {
		cout << 0;
		return 0;
	}
    vector<int> Ls(N);
    for (int i = 0; i < N; ++i)
        in >> Ls[i];

	int count = 0;
	foreach_comb(N, 3, [&](int* indexes) {
		int L0 = Ls[indexes[0]];
		int L1 = Ls[indexes[1]];
		int L2 = Ls[indexes[2]];
		if (L0 == L1 || L0 == L2 || L1 == L2)
			return;
		if ((L0 < (L1 + L2)) && (L1 < (L2 + L0)) && (L2 < (L0 + L1))) {
			++count;
		}
	});
	cout << count;

	return 0;
}