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: Accepted)

import java.io.*
import java.util.*

var pw = PrintWriter(System.out)
const val m = 31
fun solve(X: Int){
    var lis = mutableListOf<Int>(1)
    for (i in 2..m){
        var num = i * i
        while(num<=1000){
            lis.add(num)
            num *= i
        }
    }
    println(lis.sorted().findLast{ it <= X }!!)
    return
}


fun main(args: Array<String>) {
    fun StringArray(size: Int, init: (Int) -> String = { "\u0000" }): Array<String> {
        return Array<String>(size, init)
    }
    class Scanner(stream: InputStream) {
        private val reader = BufferedInputStream(stream).bufferedReader()
        private var st: StringTokenizer? = null
        fun next(): String {
            while (!(st?.hasMoreTokens() ?: false)) st = StringTokenizer(reader.readLine())
            return st!!.nextToken()
        }
    }
    val sc = Scanner(System.`in`)
    val X = sc.next().toInt()
    solve(X)
}