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)

#include <bits/stdc++.h>
using namespace std;

#define INF_LL (int64)1e18
#define INF (int32)1e9
#define REP(i, n) for(int64 i = 0;i < (n);i++)
#define FOR(i, a, b) for(int64 i = (a);i < (b);i++)
#define all(x) x.begin(),x.end()
#define fs first
#define sc second

using int32 = int_fast32_t;
using uint32 = uint_fast32_t;
using int64 = int_fast64_t;
using uint64 = uint_fast64_t;
using PII = pair<int32, int32>;
using PLL = pair<int64, int64>;

const double eps = 1e-10;

template<typename A, typename B>inline void chmin(A &a, B b){if(a > b) a = b;}
template<typename A, typename B>inline void chmax(A &a, B b){if(a < b) a = b;}


int main(void){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int64 N;
	cin >> N;
	string s="";
	while(N){
		if(N%2) s = "1"+s;
		else s = "0"+s;
		if(N%2){
			N--;
		}
		N = N/-2;
	}
	if(s.size() == 0) s = "0";
	cout << s << endl;
}