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 Java (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.ArrayList;
import java.util.Collection;
import java.util.Collections;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Object o = run();
        if (o != null)
            System.out.println(o);
    }

    static Object run() {
        try (Scanner sc = new Scanner(System.in)) {
            long n = Math.abs(sc.nextLong());
            ArrayList<Integer> l = new ArrayList<>();
            for (int i = 0; i < n; i++)
                l.add(sc.nextInt());

            long result = 0;

            for (int i = 0; i < n - 2; i++) {
                for (int j = i + 1; j < n - 1; j++) {
                    for (int k = j + 1; k < n; k++) {
                        if (l.get(i) == l.get(j) || l.get(j) == l.get(k) || l.get(k) == l.get(i))
                            continue;
                        if (l.get(i) + l.get(j) <= l.get(k) || l.get(j) + l.get(k) <= l.get(i)
                                || l.get(k) + l.get(i) <= l.get(j))
                            continue;
                        result++;
                    }
                }
            }
            return result;
        }
    }
}

class ChenNumberUtil {
    static Collection<Integer> getDivisors(int n) {
        ArrayList<Integer> al = new ArrayList<>();
        for (int i = 1; i * i <= n; i++) {
            if (n % i == 0) {
                al.add(i);
                if (n / i != i)
                    al.add(n / i);
            }
        }
        Collections.sort(al);
        return al;
    }
}

class ChenBitUtil {
    static int popCount(int bit) {
        return Integer.bitCount(bit);
    }

    static boolean isFlag(int bit, int i) {
        return 0 != (bit & (1 << i));
    }

    static int setTrue(int bit, int i) {
        return bit |= (1 << i);
    }

    static int setFalse(int bit, int i) {
        return bit &= ~(1 << i);
    }

    static int setReverse(int bit, int i) {
        return isFlag(bit, i) ? setFalse(bit, i) : setTrue(bit, i);
    }

    static int reverseAll(int bit) {
        return ~bit;
    }
}