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: Wrong Answer)

use std::io::prelude::*;

fn input<T: std::str::FromStr>() -> T {
    let stdin = std::io::stdin();
    let token: String = stdin
        .lock()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().unwrap()
}

#[allow(dead_code)]
fn with_stdout_lock<F: FnOnce(std::io::BufWriter<std::io::StdoutLock>)>(f: F) {
    let out = std::io::stdout();
    let w = std::io::BufWriter::new(out.lock());
    f(w);
}

#[allow(unused_macros)]
macro_rules! debug {
    ($var:expr) => {
        match &$var { v => eprintln!(concat!(stringify!($var), "={:?}"), v) }
    };
    ($var:expr, $($vars:expr),*) => {
        match &$var { v => eprint!(concat!(stringify!($var), "={:?} "), v) }
        debug!($($vars),*);
    };
}

fn main() {
    let n: i32 = input();

    if n == 0 {
        println!("0");
    } else if n > 0 {
        let mut m = 0;
        let mut k = 0;
        while m < n {
            m += 1 << k as i32;
            k += 2;
        }
        let mut v = vec![0; k - 1];
        v[0] = 1;
        let diff = m - n;
        for i in 1..k - 1 {
            let j = k - 1 - i - 1;
            if ((diff >> j as i32) & 1 == 1 && j & 1 == 1)
                || ((diff >> j as i32) & 1 == 0 && j & 1 == 0)
            {
                v[i] = 1;
            }
        }
        with_stdout_lock(|mut w| {
            for v in v {
                write!(w, "{}", v).unwrap();
            }
            writeln!(w, "").unwrap();
        });
    } else {
        let mut m = 0;
        let mut k = 1;
        while m > n {
            m -= 1 << k as i32;
            k += 2;
        }
        let mut v = vec![0; k - 1];
        v[0] = 1;
        let diff = m.abs() - n.abs();
        for i in 1..k - 1 {
            let j = k - 1 - i - 1;
            if ((diff >> j as i32) & 1 == 1 && j & 1 == 0)
                || ((diff >> j as i32) & 1 == 0 && j & 1 == 1)
            {
                v[i] = 1;
            }
        }
        with_stdout_lock(|mut w| {
            for v in v {
                write!(w, "{}", v).unwrap();
            }
            writeln!(w, "").unwrap();
        });
    }
}