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 C# (.NET Core 3.1.201) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;

class Program
{
    static void Main(string[] args)
    {
        var x = long.Parse(Console.ReadLine());

        long b = 0;
        long a = 0;

        if(x % 2 == 0)
        {
            double last = x;
            // a,b == 偶・偶数
            // 割れることを狙えば間に合う？
            for (long i = 0; i < x; i = i + 2)
            {
                b = 0;
                a = i;
                var tmp = Math.Pow(a, 5) - Math.Pow(b, 5);
                var c = x - tmp;
                if (tmp == x) goto ans;
                if (last < c) break;
                last = c;
            }

            for (long i = 2; i < x; i = i + 2)
            {
                b = 2;
                a = i;
                var tmp = Math.Pow(a, 5) - Math.Pow(b, 5);
                var c = x - tmp;
                if (tmp == x) goto ans;
                if (last < c) break;
                last = c;
            }
        }
        else
        {
            double last = x;
            // a,b => 偶・奇
            // 割れることを狙えば間に合う？
            for (long i = 0; i < x; i = i + 2)
            {
                b = -1;
                a = i;
                var tmp = Math.Pow(a, 5) - Math.Pow(b, 5);
                var c = x - tmp;
                if (tmp == x) goto ans;
                if (last < c) break;
                last = c;
            }
            // a,b => 偶・奇
            // 割れることを狙えば間に合う？
            for (long i = 0; i < x; i = i + 2)
            {
                b = 3;
                a = i;
                var tmp = Math.Pow(a, 5) - Math.Pow(b, 5);
                var c = x - tmp;
                if (tmp == x) goto ans;
                if (last < c) break;
                last = c;
            }
        }
        

    ans:
            Console.WriteLine($"{a} {b}");
    }
}