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 Kotlin (1.3.71) (judge verdict: Accepted)


fun main() {
    b()
}

fun a() {
    val s = readLine()!!
    fun aa(): Int {
        if (s == "RRR") return 3
        if (s.take(2) == "RR" || s.takeLast(2) == "RR") return 2
        if (s.any { it == 'R' }) return 1
        return 0
    }
    aa().toLong().print()
}

fun b() {
    val N = readLong()
    val L = readLongList().sorted()
    var ans = 0L
    L.forEachIndexed { i, l ->
        L.drop(i + 1).forEachIndexed{ j, m ->
            L.drop(i + j + 2).forEachIndexed { k, n ->
                if (l + m > n) {
                    if (l != m && l != n && m != n) {
                        ans++
                    }
                }
            }
        }
    }
    ans.print()
}

fun readLong() = readLine()!!.toLong()
fun readLongList() = readLine()!!.split(" ").map { it.toLong() }

fun Long.bitEach(action: (Int) -> Unit) {
    (0 until Long.SIZE_BITS).forEach { action(((this shr it) and 1).toInt()) }
}

fun Long.bitFilter(bit: Int) = (0 until Long.SIZE_BITS).filter { ((this shr it) and 1) == 1L }

fun <T1, T2> pairs(first: Iterable<T1>, second: Iterable<T2>) = first.flatMap { t1 -> second.map { t2 -> t1 to t2 } }

fun Long.primeSequence() = sequence<Long> {
    var n = this@primeSequence
    if (n % 2L == 0L) {
        yield(2L)
        n /= 2L
    }
    var i = 3L
    while (n > 0) {
        if (n % i == 0L) {
            yield(i)
            n /= i
        }
        i += 2L
    }
}

fun Long.print() {
    println(this)
}