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)

#![allow(unused_imports)]
use std::io::*;
use std::fmt::*;
use std::str::*;
use std::cmp::*;
use std::collections::*;

pub trait InputValue {
    fn parse(s: &str) -> Self;
}

pub fn read<T: InputValue>() -> T {
    let mut buf = String::new();
    let _ = stdin().read_line(&mut buf);
    T::parse(&buf.trim())
}

pub fn readn<T: InputValue>(n: usize) -> Vec<T> {
    let mut vec = vec![];
    for _ in 0..n {
        vec.push(read());
    }
    vec
}

macro_rules! parse_single_value {
    ($($t:ty),*) => {
        $(
            impl InputValue for $t {
                fn parse(s: &str) -> $t { s.parse().unwrap() }
            }
        )*
	}
}
parse_single_value!(i32, i64, f32, f64, usize, String);

macro_rules! parse_tuple {
	($($t:ident),*) => {
		impl<$($t),*> InputValue for ($($t),*) where $($t: InputValue),* {
			fn parse(s: &str) -> ($($t),*) {
				let mut tokens = s.split_whitespace();
				let t = ($($t::parse(tokens.next().unwrap())),*);
				t
			}
		}
	}
}
parse_tuple!(A, B);
parse_tuple!(A, B, C);

// ===

fn main() {
    let n: i64 = read();
    if n == 0 {
        println!("{}", 0);
    } else {
        let mut powm2 = vec![0; 48];
        for i in 0..48 {
            powm2[i] = 1<<i;
            if i % 2 == 1 {
                powm2[i] *= -1;
            }
        }

        let mut ans: Vec<i32> = vec![0; 48];
        let abs = n.abs();
        let mut sum: i64 = 0;
        let mut idx = 0;
        if n < 0 {
            idx = 1;
        }

        {
            while sum.abs() < abs {
                sum += powm2[idx];
                ans[idx] = 1;
                idx += 2;
            }
        }
        {
            loop {
                if sum.abs() - (powm2[idx]).abs() >= n.abs() {
                    if ans[idx] == 1 {
                        sum -= powm2[idx];
                    } else {
                        sum += powm2[idx];
                    }
                    ans[idx] ^= 1;
                }
                if idx == 0 {
                    break;
                }
                idx -= 1;
            }
        }
        let mut nonzero = 0;
        for i in (0..48).rev() {
            if ans[i] == 1 {
                nonzero = 1;
            }
            if nonzero == 1 {
                if ans[i] == 0 {
                    print!("0")
                } else {
                    print!("1")
                }
            }
        }
        println!();
    }


    // 1, (-2), 4, (-8), 16, (-32), 64, (-128),...
    // 1   0    1   1    1         = 13
    // 0   1    1   0    1     1   = -13
}