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 __USE_MINGW_ANSI_STDIO 0
#include <iostream>
#include <string>
#include <vector>
#include <stdio.h>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <set>
#include <numeric>
#include <bitset>
#include <map>
#include <limits.h>
using namespace std;
typedef unsigned long ul;
typedef pair<ul, ul> P;
const int INF = 1e9;

long long gcd(long long x, long long y){
			if(x<y){
				long long temp = x;
			x = y;
				y = temp;
			}
			long long r = x % y;
				while(r!=0){
					x = y;
					y = r;
				r = x % y;
	    		}
			return y;
	    }

int main(){
   long long N;
   cin >> N;
   string ans;
   while(N != 0){
       int r = N % (-2);
       if(r == -1)r = 1;
       N = (N-r) / (-2);
    ans.push_back((char)('0'+ r));
   }
reverse(ans.begin(),ans.end());
if(ans == "")ans = "0";
cout << ans << endl;
}