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 (1.15.1) (judge verdict: Accepted)

use std::str::FromStr;
use std::fmt::Debug;
use std::cmp;

#[allow(dead_code)]
fn read_as_vec<T>() -> Vec<T>
where T: FromStr, <T as FromStr>::Err : Debug{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().split_whitespace().map(|c| T::from_str(c).unwrap()).collect()

}

fn main(){
    let mut n = read_as_vec::<i64>()[0];
    let mut bit: i64 = 1;
    let mut res: Vec<bool> = vec![];

    if n == 0{
        println!("0");
        return;
    }

    while n != 0{
        if n%(bit*-2) == 0{

            res.push(false);
        }else{
            res.push(true);
            n -= bit;
        }
        bit *= -2;
    }

    for b in res.into_iter().rev(){
        if b{
            print!("1");
        }else{
            print!("0");
        }
    }
    println!("");
}