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 Ruby (2.7.1) (judge verdict: Runtime Error)

# -2進数でn桁のときの取りうる最大値と最小値を返す
def cal_min_max n
  if n.even?
    max = (0..n).step(2).inject(0){|sum, i| sum + (-2) ** i}
    min = (1..n).step(2).inject(0){|sum, i| sum + (-2) ** i} + (-2) ** n
  else
    max = (0..n).step(2).inject(0){|sum, i| sum + (-2) ** i} + (-2) ** n
    min = (1..n).step(2).inject(0){|sum, i| sum + (-2) ** i}
  end
  [max, min]
end

n = gets.chomp.to_i
ans_hash = Hash.new{nil}

until n.zero? do
  i = Math.log(n.abs, 2).to_i
  i = if n > 0
    i.to_i.even? ? i : i + 1
  else # n < 0
    i.to_i.odd? ? i : i + 1
  end
  max, min = cal_min_max(i)
  i += 2 unless min <= n && n <= max
  ans_hash[i] = 1
  n -= (-2) ** i
end

digits = ans_hash.keys.max
ans = "0" * (digits + 1)
ans_hash.each do |k, _v|
  ans[k] = "1"
end

puts ans.reverse