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 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
#include <math.h>
#include <string.h>

int main() {
	long long int N, i, n = 0, j;
	char S[64];
	memset(S, '0', sizeof(S));

	scanf("%lld", &N);

	if (N == 0) {
		printf("0");
		return 0;
	}
	else if (N > 0) {
		for (i = 0; n < N; i += 2) {
			n = n + (int)pow(2.0, (double)i);
			S[i] = '1';
		}
		i -= 2;
		for (j = 1; j <= i; j++) {
			if (N <= n - (int)pow(2.0, (double)(i - j))) {
				n = n - (int)pow(2.0, (double)(i - j));
				if (j % 2 == 1) {
					S[j] = '1';
				}
				else {
					S[j] = '0';
				}
			}
		}
	}
	else {
		for (i = 1; n > N; i += 2) {
			n = n - (int)pow(2.0, (double)i);
			S[i - 1] = '1';
		}
		i -= 2;
		for (j = 1; j <= i; j++) {
			if (N >= n + (int)pow(2.0, (double)(i - j))) {
				n = n + (int)pow(2.0, (double)(i - j));
				if (j % 2 == 1) {
					S[j] = '1';
				}
				else {
					S[j] = '0';
				}
			}
		}
	}

	S[i + 1] = '\0';
	printf("%s", S);

	return 0;
}