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: Accepted)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Collections.Specialized;
using System.Linq;
using System.Numerics;

namespace AtCoderWorkspace
{
    public class Solver
    {
        public void Solve()
        {
            var cin = new Scanner();
            var x = cin.nextInt();

            var ll = Enumerable.Range(-200, 400).ToList();
            for (int i = 0; i < 400; i++)
            {
                for (int j = 0; j < 400; j++)
                {
                    var ans = Math.Pow(ll[i], 5) - Math.Pow(ll[j], 5);
                    if (ans == x)
                    {
                        Console.WriteLine(ll[i] + " " + ll[j]);
                        return;
                    }
                }
            }            
        }
    }

    // 組み合わせ列挙 withRepetition = true (重複組み合わせ) 
    public static class Combination
    {
        public static IEnumerable<T[]> Enumerate<T>(IEnumerable<T> items, int k, bool withRepetition)
        {
            if (k == 1)
            {
                foreach (var item in items)
                    yield return new T[] { item };
                yield break;
            }
            foreach (var item in items)
            {
                var leftside = new T[] { item };

                var unused = withRepetition ? items : items.SkipWhile(e => !e.Equals(item)).Skip(1).ToList();

                foreach (var rightside in Enumerate(unused, k - 1, withRepetition))
                {
                    yield return leftside.Concat(rightside).ToArray();
                }
            }
        }
    }

    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 int[] DPArrayInt(int N, int ex)
        {
            int[] Array = new int[N + ex];
            for (int i = 0; i < N; i++)
            {
                Array[i] = nextInt();
            }
            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 long[] DPArrayLong(int N, int ex)
        {
            long[] Array = new long[N + ex];
            for (int i = 0; i < N; i++)
            {
                Array[i] = nextLong();
            }
            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;
        }
    }

    class Program
    {
        static void Main(string[] s) => new Solver().Solve();
    }
}