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++14 (GCC 5.4.1) (judge verdict: Accepted)

#define _CRT_SECURE_NO_WARNINGS
#include <algorithm>
#include<cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include<iostream>
#include<list>
#include<queue>
#include<set>
#include<string>
#include<utility>
#include <vector>

#define ll long long
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)   FOR(i,0,n)
#define FORR(i,a,b) for(int i=(a);i>=(b);--i)
#define REPR(i,n)   FORR(i,n,0)

using namespace std;

int main(){
	int N, key = 0;
	char c[100];

	scanf("%d", &N);
	if (N == 0){
		printf("0\n");
		return 0;
	}
	else{
		while (N != 1){
			if (abs(N) % 2 == 1){
				c[key] = '1';
			}
			else{
				c[key] = '0';
			}
			if (N < 0 && abs(N) % 2 == 1) {
				N = abs(N / 2) + 1;
			}
			else{
				N = -1 * N / 2;
			}
			key++;
		}
		c[key] = '1';
	}

	REPR(i, key){
		printf("%c", c[i]);
	}
	printf("\n");

	return 0;
}