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: Accepted)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <stdint.h> // uint64_t

#define max(a,b) ((a) > (b) ? (a) : (b))
#define min(a,b) ((a) > (b) ? (b) : (a))

int get_int(void) {
  int num;
  scanf("%d", &num);
  return num;
}

#define POWER2_MAX 32

int main(void) {
    int num = get_int();
    int sign = (num>=0) ? 1 : -1;
    int anum = abs(num);
    static int map[POWER2_MAX];
    int bit;
    for(bit = 0; bit < POWER2_MAX; bit++) {
        if(!(anum & (1<<bit))) continue;
        int bit_sign = (bit%2) ? -1 : 1;
        map[bit] = bit_sign * sign;
    }

    // carry
    int msb = 0;
    for(bit = 0; bit < POWER2_MAX; bit++) {
        if(map[bit]==-1) {
            map[bit] = 1;
            map[bit+1] += 1;
        } else if (map[bit]==2) {
            map[bit] = 0;
            map[bit+1] -= 1;
        } else if (map[bit] == -2) {
            map[bit] = 0;
            map[bit+1] += 1;
        }
        if(map[bit]==1) msb = bit;
    }

    for(bit = msb; bit >= 0; bit--) {
        putchar('0' + map[bit]);
    }
    putchar('\n');
    return 0;
}