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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

int power(int a, int N){
	if(N == 0){
		return 1;
	}
	else if(N % 2 == 0){
		return power(a * a, N / 2);
	}
	else{
		return a * power(a, N - 1);
	}
}

int ABS(int a){
	return a >= 0 ? a : -a;
}

int main(){
	int N, i, j, f = 0, inf, sup, v;
	scanf("%d", &N);
	if(N == 0){
		printf("0\n");
		return 0;
	}
	for(i = 30; i >= 0; i--){
		v = power(i % 2 == 0 ? 2 : -2, i);
		sup = v;
		inf = v;
		for(j = i - 1; j >= 0; j--){
			if(j % 2 == 0){
				sup += power(2, j);
			}
			else{
				inf += power(-2, j);
			}
		}
		if(inf <= N && N <= sup){
			printf("1");
			N -= v;
			f = 1;
		}
		else{
			if(f == 1){
				printf("0");
			}
		}
	}
	printf("\n");
	return 0;
}