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: Wrong Answer)



import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		
		//  1 <= x <= 1000
		int x = scanner.nextInt();
		int ans = 0;
		ans = exponential(x);
		System.out.println(ans);
		
	}
	
	public static int exponential(int x) {
		int ans = 0;
		for (int i = 31; i >= 1; i--) {
			ans = 0;
			for (int j = 9; j >= 2; j--) {
				if (Math.pow((double)i, (double)j) <= x) {
					ans = Math.max(ans, (int)Math.pow((double)i, (double)j));
					return ans;
				}
			}
		}
		return ans;
	}

}