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 (2.3.3) (judge verdict: Wrong Answer)

N = -9

puts 0 if N.zero?

# +: 1 から 4 倍にしていく 1, 4, 16, 64
def plus(n)
  raise if n <= 0
  index = 0
  b = 1
  while true
    return [index, n-b] if n < b * 4
    index += 2
    b *= 4
  end
end

# -: -2 から 4 倍にしていく -2, -8, -32
def minus(n)
  raise if n >= 0
  index = 1
  b = -2
  while true
    return [index, n-b] if n > b * 4
    index += 2
    b *= 4
  end
end

def f(n, a=[0])
  if n.zero?
    puts a.map { |n| n.to_i.to_s }.reverse.join
    exit
  elsif n.negative?
    index, next_n = minus(n)
    a[index] = 1
    f(next_n, a)
  else
    index, next_n = plus(n)
    a[index] = 1
    f(next_n, a)
  end
end

f(N)

# 00001 -> 1
# 00010 -> -2
# 00100 -> 4
# 01000 -> -8
# 10000 -> 16

# 00001 -> 1
# 00010 -> -2
# 00011 -> -1
# 00100 -> 4
# 00101 -> 5
# 00110 -> 2
# 00111 -> 3
# 01000 -> -8
# 01001 -> -7
# 01010 -> -10
# 01011 -> -9
# 01100 -> -4
# 01101 -> -3
# 01110 -> -6
# 01111 -> -5
# 10000 -> 16