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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.*;

class Main
{
    public static void main(String[] args) throws IOException
    {
	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	
	String str = br.readLine();

	int X = Integer.parseInt(str);

	find_max(2, 2, X, 1);
    }

    public static void find_max(int e, int x, int X, int max)
    {
	if(X == 1 || X == 2 || X == 3){
	    System.out.println(1);
	}else{
	    if(Math.pow((double) e, (double) x) <= (double) X){
		if(Math.pow((double) e, (double) x) > (double) max){
		    find_max(e, x+1, X, (int) Math.pow((double) e, (double) x));
		}else{
		    find_max(e, x+1, X, max);
		}
	    }else{
		if(Math.pow((double) (e+1), (double) 2) <= (double) X){
		    find_max(e+1, 2, X, max);
		}else{
		    System.out.println(max);
		}
	    }
	}
    }
}