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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)



import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Main {

	static Scanner scanner;

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

        int X=gi();
        
        int max=1;
        for(int i=2; i<Math.floorDiv(X, 2)+1; i++) {
        	int z=2;
        	while(true) {
        		int v=(int)Math.pow(i,z);
        		if(v>X) {
        			break;
        		}
        		if(max<v) {
        			max=v;
        		}
        		z++;
        	}
        }


	    System.out.println(max);

//	    if (isNumber(s1) && isNumber(s2) && s.charAt(a) == '-') {
//	    	System.out.print("Yes");
//	    }else {
//	    	System.out.print("No");
//	    }



	}

	public static long d(long n) {
		String a=String.valueOf(n);
		return (long)a.length();
	}

	// 文字列として入力を取得
	public static String gs() {
		return scanner.next();
	}

	// intとして入力を取得
	public static int gi() {
		return Integer.parseInt(scanner.next());
	}

	// longとして入力を取得
	public static long gl() {
		return Long.parseLong(scanner.next());
	}

	// doubleとして入力を取得
	public static double gd() {
		return Double.parseDouble(scanner.next());
	}
}