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: Wrong Answer)

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

int main()
{
	long long n, i, j, pivot, d, s;
	scanf("%lld", &n);

	s = 0;
	for (i = 32; i >= 1; i --) {
		pivot = 0;
		for (j = 1; j < i; j ++) {
			pivot -= (long long)pow(-2, j) - (long long)pow(-1, j);
		}

		if (i % 2 == 0) {
			if (pivot >= n) {
				d = (pivot - n) % (long long)pow(2,i) >= (long long)pow(2,i-1) ? 1 : 0;
			} else {
				d = (n - (pivot + 1)) % (long long)pow(2,i) >= (long long)pow(2,i-1) ? 0 : 1;
			}
		} else {
			if (pivot <= n) {
				d = (n - pivot) % (long long)pow(2,i) >= (long long)pow(2,i-1) ? 1 : 0;
			} else {
				d = ((pivot - 1) - n) % (long long)pow(2,i) >= (long long)pow(2,i-1) ? 0 : 1;
			}
		}

		if (d == 1) {
			s = 1;
		}

		if (s == 1) {
			printf ("%d", d);
		}
	}
	return 0;
}