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 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
#include <math.h>
int main(void){
    // Your code here!
    int N;
    scanf ("%d",&N);
    int i;
    char bits[32];
    long sum = 0;
    int maxbit = 0;
    for ( i = 0 ; 1 ; i ++) //終了条件に意味なし
    {
        if ( (N - sum) % (1 << (i + 1))  == 0 )
        {
            bits[i] = 0;
            //printf("%dketameha0\n",i);
        }
        else
        {
            //printf("%dketameha1\n",i);
            bits[i] = 1;
            if ( i % 2 == 0 )
            {
                sum += (1 << i);
            }
            else
            {
                sum -= (1 << i);
            }
        }
        
        if ( sum == N)
            break;
        
        maxbit ++;
    }
    
    for (i = maxbit ; i >= 0 ; i -- )
    {
        printf("%s", bits[i] ? "1" : "0");
    }
    printf("\n");
    return 0;
    
}