Problem:
Score : 200 points

Problem Statement

You are given a positive integer X.
Find the largest perfect power that is at most X.
Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.

Constraints

1 ≤ X ≤ 1000

X is an integer.

Input

Input is given from Standard Input in the following format:

X

Output

Print the largest perfect power that is at most X.

Sample Input 1

10

Sample Output 1

9

There are four perfect powers that are at most 10: 1, 4, 8 and 9.
We should print the largest among them, 9.

Sample Input 2

1

Sample Output 2

1

Sample Input 3

999

Sample Output 3

961

Solution in Kotlin (1.0.0) (judge verdict: Wrong Answer)

fun main(args: Array<String>) {

    val x = readInts()[0]

    println(getList().maxBy { it <= x })
}

fun getList() =
    (1 .. 31).flatMap {base ->
        (1 .. 9).map { Math.pow(base.toDouble(), it.toDouble()).toInt() }.filter { it <= 1000 }
    }.distinct().sortedDescending()

fun readInts(): List<Int> = readStrings().map { it.toInt() }
fun readStrings(): List<String> = readLine()!!.split(" ")