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;

//const long double PIL = 3.141592653589793238462643383279502884L;
//const double PI = 3.14159265358979323846;

#define sz(a) int((a).size())
#define all(c) (c).begin(), (c).end()
#define what_is(x) cerr << #x << " is " << x << endl;

using ll = long long;
using ii = pair<int,int>;
using vi = vector<int>;
using vll = vector<long long>;
using vii = vector<pair<int,int>>;

int main() {
	//~ ios::sync_with_stdio(0);
	//~ cin.tie(0);
	int n;
	cin >> n;
	if (n == 0) {
		puts("0");
		return 0;
	}

	vi v;
	for (int i = 0; n != 0; i = 1 - i) {
		v.push_back(n&1);
		if (n&1) {
			if (n > 0 && i == 1) n += 2;
			if (n < 0 && i == 0) n -= 2;
		}
		n /= 2;
	}
	for (int i = v.size() - 1; i >= 0; i--) cout << v[i];
	cout << endl;
}