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 Java7 (OpenJDK 1.7.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

import javax.sql.rowset.serial.SerialArray;

public class Main {

		public static void main(String[] args){
			
			Scanner sc = new Scanner(System.in);

			
			int X = sc.nextInt();

			int ans = 0;
			int max = 0;
			double d = Math.sqrt(X);
			
			int wki = (int) Math.floor(d);
			
			for(int i = 2;i <= wki;i++){
				for(int j = 2;j <= wki;j++){
					ans = (int) Math.pow(i, j);
				    System.out.println(i);
				    System.out.println(j);
				    System.out.println(ans);
					if(X <= ans ) {
						j = wki;
					}else{
						if(max < ans){
							max = ans;
						}
					}
				}
			}
			
			if (max == 0){
			    System.out.println(X);
			    return ;
				
			}
		    System.out.println(max);
		    return ;
		}
			
}