Problem:
Score : 400 points

Problem Statement

Give a pair of integers (A, B) such that A^5-B^5 = X.
It is guaranteed that there exists such a pair for the given integer X.

Constraints

1 \leq X \leq 10^9

X is an integer.

There exists a pair of integers (A, B) satisfying the condition in Problem Statement.

Input

Input is given from Standard Input in the following format:

X

Output

Print A and B, with space in between.
If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them.

A B

Sample Input 1

33

Sample Output 1

2 -1

For A=2 and B=-1, A^5-B^5 = 33.

Sample Input 2

1

Sample Output 2

0 -1

Solution in Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.io.IOException;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) throws IOException {
		//		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		//		String W	 = in.readLine();
		//		int num = Integer.parseInt(W.split(" ")[0]);

		Scanner sc = new Scanner(System.in);
		final long X = sc.nextInt();
		Map<Integer, Long> fivePow = new HashMap<Integer, Long>();
		long tmp = 0;

		long max = (long) (Math.pow(10, 10));

		for (int i = 0; i == 0 || tmp < max; i++) {
			tmp = (long) Math.pow(i, 5);
			fivePow.put(i, tmp);
		}

		for (int i = 0; i == 0 || tmp > -max; i--) {
			tmp = (long) Math.pow(i, 5);
			fivePow.put(i, tmp);
		}

		long A = 0;
		long B = 0;

		for (Integer a : fivePow.keySet()) {
			tmp = X + fivePow.get(a);
			if (fivePow.containsValue(tmp)) {
				B = a;
				A = (long) Math.pow(tmp, 0.2);
				break;
			}
		}

		System.out.println(A + " " + B);
	}

	//	//二分探索
	//k <= num となる最小の配列要素kのインデックスを返す
	static private int binarySearch(long num, long[] orderedArray) {
		int lowerBorder = -1;
		int upperBorder = orderedArray.length;
		int mid;

		while (upperBorder - lowerBorder > 1) {
			mid = (upperBorder + lowerBorder) / 2;
			if (orderedArray[mid] <= num) {
				lowerBorder = mid;
			} else {
				upperBorder = mid;
			}
		}
		return lowerBorder;
	}

	//二分探索
	//k <= num となる最小のList要素kのインデックスを返す
	static private int binarySearch(long num, ArrayList<Long> orderedList) {
		int lowerBorder = -1;
		int upperBorder = orderedList.size();
		int mid;

		while (upperBorder - lowerBorder > 1) {
			mid = (upperBorder + lowerBorder) / 2;
			if (orderedList.get(mid) <= num) {
				lowerBorder = mid;
			} else {
				upperBorder = mid;
			}
		}
		return lowerBorder;
	}

	//aとbの最小公倍数を求める
	public static int gcd(int a, int b) {
		return b == 0 ? a : gcd(b, a % b);
	}

	public static long gcd(long a, long b) {
		return b == 0 ? a : gcd(b, a % b);
	}

}