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;
using System.Text;
using System.Threading.Tasks;
using static System.Console;

class Program
{
    static void Main(string[] args)
    {
        new Program().Calc();
    }

    Scanner cin;

    public Program() { }

    bool chmax<T>(ref T a, T b) where T : IComparable<T> { if (a.CompareTo(b) < 0) { a = b; return true; } return false; }
    bool chmin<T>(ref T a, T b) where T : IComparable<T> { if (b.CompareTo(a) < 0) { a = b; return true; } return false; }

    void Calc()
    {
        cin = new Scanner();

        long x = cin.nextLong();
        int a, b = 0;
        bool found = false;
        for (a = 0; a < 100; ++a)
        {
            for (b = -100; b < 100; ++b)
            {
                int ans = a * a * a * a * a - b * b * b * b * b;
                if (ans == x)
                {
                    found = true;
                    break;
                }
            }

            if (found) break;
        }

        WriteLine(a.ToString() + " " + b.ToString());
    }

    private double NthRoot(double x, int n)
    {
        if (x >= 0)
        {
            // 通常のn乗根
            return Math.Pow(x, 1.0 / (double)n);
        }
        else
        {
            if ((n % 2) != 0)
            {
                // 奇数乗根の場合
                return -Math.Pow(-x, 1.0 / (double)n);
            }
            else
            {
                // 偶数乗根の場合
                return double.NaN;
            }
        }
    }
}

public static class StringExtension
{
    public static char GetChar1(this string str, int indexOneBase)
    {
        return str[indexOneBase - 1];
    }
}

class Scanner
{
    string[] s;
    int i;

    char[] cs = new char[] { ' ' };

    public Scanner()
    {
        s = new string[0];
        i = 0;
    }

    public string next()
    {
        if (i < s.Length) return s[i++];
        string st = Console.ReadLine();
        while (st == "") st = Console.ReadLine();
        s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
        if (s.Length == 0) return next();
        i = 0;
        return s[i++];
    }

    public int nextInt()
    {
        return int.Parse(next());
    }
    public int[] ArrayInt(int N, int add = 0)
    {
        int[] Array = new int[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextInt() + add;
        }
        return Array;
    }

    public long nextLong()
    {
        return long.Parse(next());
    }

    public long[] ArrayLong(int N, long add = 0)
    {
        long[] Array = new long[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextLong() + add;
        }
        return Array;
    }

    public double nextDouble()
    {
        return double.Parse(next());
    }


    public double[] ArrayDouble(int N, double add = 0)
    {
        double[] Array = new double[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextDouble() + add;
        }
        return Array;
    }
}