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.Generic;
using System.Diagnostics;
using System.Linq;
using System.Runtime.CompilerServices;
using System.Text;

namespace AtCoder
{
    static class Program_D
    {
        class Solver
        {
            
            public void Solve()
            {
                var x = Int();
                Int64 max=1024;
                for(Int64 a=0; a<max;++a)
                {
                    for(Int64 b=-max; b<max; ++b)
                    {
                        Int64 calc = a*a*a * a * a - b * b * b * b * b;
                        if(calc == x)
                        {
                            Console.WriteLine($"{a} {b}");
                            return;
                        }
                    }
                }


            }




            // ----------------------------------------------------------------------------

            public int Int()
            {
                return int.Parse(Console.ReadLine());
            }
            public int[] Ints(int subtract)
            {
                return Console.ReadLine().Split(' ').Select(s => int.Parse(s) - subtract).ToArray();
            }
            public long[] Longs()
            {
                return Console.ReadLine().Split(' ').Select(s => long.Parse(s)).ToArray();
            }
        }



        static void Main(string[] args)
        {
            new Solver().Solve();
        }
    }
    struct ModInt
    {
        const int MOD = 1000000007;
        long Data;
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public ModInt(long data) { if ((Data = data % MOD) < 0) Data += MOD; }
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static implicit operator long(ModInt modInt) => modInt.Data;
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static implicit operator ModInt(long val) => new ModInt(val);
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static ModInt operator +(ModInt a, ModInt b)
        {
            long res = a.Data + b.Data;
            return new ModInt() { Data = res >= MOD ? res - MOD : res };
        }
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static ModInt operator +(ModInt a, long b) => a.Data + b;
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static ModInt operator -(ModInt a, ModInt b)
        {
            long res = a.Data - b.Data;
            return new ModInt() { Data = res < 0 ? res + MOD : res };
        }
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static ModInt operator -(ModInt a, long b) => a.Data - b;
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static ModInt operator *(ModInt a, ModInt b) => new ModInt() { Data = a.Data * b.Data % MOD };
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public static ModInt operator /(ModInt a, ModInt b) => a.Data * GetInverse(b);

        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        public override string ToString() => Data.ToString();

        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        static long GetInverse(long a)
        {
            long div;
            long p = MOD;
            long x1 = 1, y1 = 0, x2 = 0, y2 = 1;
            while (true)
            {
                if (p == 1) return x2;
                div = a / p;
                x1 -= x2 * div;
                y1 -= y2 * div;
                a %= p;
                if (a == 1) return x1;
                div = p / a;
                x2 -= x1 * div;
                y2 -= y1 * div;
                p %= a;
            }
        }
    }

}