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.*;
public class Main {

		static void solve() {
			Scanner sc = new Scanner(System.in);
			int n = sc.nextInt();
			int min = Integer.MAX_VALUE;
			int result = 0;
			sc.close();
			for(int b = 1;b<32;b++) {
				for(int p = 2;p<32;p++) {
					if(Math.pow(b, p)<1000) {
					if(Math.pow(b, p)<=n) {
						if(n-Math.pow(b, p)<min) {
							min = (int) (n - Math.pow(b, p));
							result = (int) Math.pow(b, p);
						}
					}
					else if(Math.pow(b, p)>=n) {
						if(Math.pow(b, p)-n<min) {
							min = (int) (n - Math.pow(b, p));
							result = (int) Math.pow(b, p);
						}
					}
				}
			}
		}
			System.out.println(result);
		}
	public static void main(String[] args) {
		solve();

	}

}