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 <cstdio>
#include <cmath>
#include <iostream>
#include <sstream>
#include <string>
#include <vector>
#include <map>
#include <queue>
#include <algorithm>

#ifdef _DEBUG
#define DMP(x) cerr << #x << ": " << x << "\n"
#else
#define DMP(x) ((void)0)
#endif

const int MOD = 1000000007, INF = 1111111111; 
using namespace std;
typedef long long lint;

int main() {

	cin.tie(nullptr);
	ios::sync_with_stdio(false);

	lint N;
	cin >> N;

	string ans;

	if (N >= 0) {

		int st = 1;
		for (int i = 0; i <= 32; i++) {
			if (i % 2) st += (1LL << (i - 1));
			ans += int((N - st + (1LL << (i + 1))) % (1LL << (i + 1)) < (1LL << i)) + '0';
		}
		reverse(ans.begin(), ans.end());
	}
	else {

		int st = 1;
		for (int i = 0; i <= 32; i++) {
			if (i % 2) st += (1LL << (i - 1));
			ans += int((-(N-1)/2 - st + (1LL << (i + 1))) % (1LL << (i + 1)) < (1LL << i)) + '0';
		}
		reverse(ans.begin(), ans.end());
		ans += -N % 2 + '0';
	}

	for (int i = 0; i <= 32; i++) {
		if (ans[i] == '1') {
			cout << ans.substr(i) << "\n";
			break;
		}
		if (i == 32) cout << 0 << "\n";
	}

	return 0;
}