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: Wrong Answer)

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <queue>
#include <set>
#include <map>
#include <utility>
#include <cmath>
#include <stack>
using namespace std;
#define fio ios::sync_with_stdio(0),cin.tie(0);
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;



ll N, k;
stack<char> ans;
int main(){
	scanf("%lld", &N);
	while (N) {
		ans.push(abs(N/2) + '0');
		if (N > 0)
			N >>= 1;
		else
			N = (N - 1) / 2;
		N *= -1;
	}
	bool flag = 0;
	while (!ans.empty() && ans.top() == '0') ans.pop();
	while (!ans.empty()) {
		printf("%c", ans.top());
		flag = 1;
		ans.pop();
	}
	if (!flag) printf("0");

}