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 Rust (1.42.0) (judge verdict: Accepted)

#![allow(unused_imports)]
use proconio::{input, fastout};
use proconio::marker::*;

#[fastout]
fn main() {
    input! {
    mut n: i64,
    }
    if n == 0 {
        print!("0");
        return
    }
    let mut d = vec![];
    while n != 0 {
        d.push(n % -2);
        n /= -2
    }
    for i in 0..d.len() {
        if d[i] == -1 {
            d[i] = 1;
            inc(&mut d, i + 1);
        }
    }
    for i in (0..d.len()).rev() {
        print!("{}", d[i]);
    }
}

fn inc(d: &mut Vec<i64>, t: usize) {
    if d.len() <= t {
        d.push(1);
        return
    }
    if d[t] < 1 {
        d[t] += 1;
        return
    } else {
        d[t] = 0;
        inc(d, t + 1);
        inc(d, t + 2);
    }
}