Problem:
Score : 300 points

Problem Statement

Given an integer N, find the base -2 representation of N.

Here, S is the base -2 representation of N when the following are all satisfied:

S is a string consisting of 0 and 1.

Unless S = 0, the initial character of S is 1.

Let S = S_k S_{k-1} ... S_0, then S_0 \times (-2)^0 + S_1 \times (-2)^1 + ... + S_k \times (-2)^k = N.

It can be proved that, for any integer M, the base -2 representation of M is uniquely determined.

Constraints

Every value in input is integer.

-10^9 \leq N \leq 10^9

Input

Input is given from Standard Input in the following format:

N

Output

Print the base -2 representation of N.

Sample Input 1

-9

Sample Output 1

1011

As (-2)^0 + (-2)^1 + (-2)^3 = 1 + (-2) + (-8) = -9, 1011 is the base -2 representation of -9.

Sample Input 2

123456789

Sample Output 2

11000101011001101110100010101

Sample Input 3

0

Sample Output 3

0

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

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

namespace AtCoder
{
    class Program
    {
        static void Main(string[] args)
        {
            //[summary]C - Base -2 Number
            int n = int.Parse(Console.ReadLine());

            //-2進法の最大ケタ数は30
            int max = (int)Math.Pow(2, 10);

            for(int i = 0; i < max; i++)
            {
                //10進法を2進法に変換
                string s = Convert.ToString(i, 2);

                //文字列を反転(こっちを計算に使う)
                var binary = String.Join("", s.Reverse());
                
                int decimalNum = CalcDecimalNumber(binary);

                if (decimalNum == n)
                {
                    Console.WriteLine(s);
                    break;
                }
            }          
        }

        static int CalcDecimalNumber(string binary)
        {
            //-2進数表現された"1011"のような文字列から10進法の数字を逆算

            int sum = 0;

            for(int i = 0; i < binary.Length; i++)
            {
                if (binary[i] == '1')
                {
                    sum += (int)Math.Pow(-2, i);
                }                
            }

            return sum;
        }
    }
}