Problem:
Score : 400 points

Problem Statement

Let \mathrm{popcount}(n) be the number of 1s in the binary representation of n.
For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.

Let f(n) be the number of times the following operation will be done when we repeat it until n becomes 0: "replace n with the remainder when n is divided by \mathrm{popcount}(n)." (It can be proved that, under the constraints of this problem, n always becomes 0 after a finite number of operations.)

For example, when n=7, it becomes 0 after two operations, as follows:

\mathrm{popcount}(7)=3, so we divide 7 by 3 and replace it with the remainder, 1.

\mathrm{popcount}(1)=1, so we divide 1 by 1 and replace it with the remainder, 0.

You are given an integer X with N digits in binary.
For each integer i such that 1 \leq i \leq N, let X_i be what X becomes when the i-th bit from the top is inverted.
Find f(X_1), f(X_2), \ldots, f(X_N).

Constraints

1 \leq N \leq 2 \times 10^5

X is an integer with N digits in binary, possibly with leading zeros.

Input

Input is given from Standard Input in the following format:

N
X

Output

Print N lines. The i-th line should contain the value f(X_i).

Sample Input 1

3
011

Sample Output 1

2
1
1

X_1 = 7, which will change as follows: 7 \rightarrow 1 \rightarrow 0. Thus, f(7) = 2.

X_2 = 1, which will change as follows: 1 \rightarrow 0. Thus, f(1) = 1.

X_3 = 2, which will change as follows: 2 \rightarrow 0. Thus, f(2) = 1.

Sample Input 2

23
00110111001011011001110

Sample Output 2

2
1
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
3

Solution in Ruby (2.7.1) (judge verdict: Runtime Error)

INF = 10 ** 100
n = gets.to_i
x = gets.chomp

def popcount(num)
  num.to_s(2).count('1')
end

start_count = x.count('1')

plus = start_count + 1
plus_contribution = [1]
plus_last = plus_contribution.last
INF.times do |i|
  tmp = plus_contribution.first * 2 % plus if plus != 0
  # if tmp == plus_last
  #   break
  # else
    plus_contribution.unshift(tmp)
  # end
  break if i >= n - 2
end
plus_contribution_size = plus_contribution.size

minus = start_count - 1
minus_contribution = [1]
minus_last = minus_contribution.last
INF.times do |i|
  tmp = minus_contribution.first * 2 % minus if minus != 0
  # if tmp == minus_last
  #   break
  # else
    minus_contribution.unshift(tmp)
  # end
  break if i >= n - 2
end
minus_contribution_size = minus_contribution.size

plus_default = 0
minus_default = 0
x.each_char.with_index do |c, i|
  next if c == '0'
  plus_default += plus_contribution[i]
  minus_default += minus_contribution[i]
end
plus_default %= plus if plus != 0
minus_default %= minus if minus != 0
# plus_contribution.reverse!
# minus_contribution.reverse!
# p plus_contribution
# p minus_contribution
# puts plus_default
# puts minus_default

# puts plus_default
# p plus_contribution
# puts
ans = []
(n - 1).downto(0) do |i|
  if x[i] == '0'
    tmp_mod = plus_default
    # puts tmp_mod if i == n-1
    tmp_mod += plus_contribution[i]
    tmp_mod -= plus if tmp_mod >= plus
    # puts tmp_mod if i == n-1
    # puts
    tmp_mod %= plus
  else
    tmp_mod = minus_default
    tmp_mod -= minus_contribution[i]
    tmp_mod += minus if tmp_mod < 0
    tmp_mod %= minus
  end
  # puts tmp_mod if i == n-1
  count = 1
  # puts tmp_mod
  until tmp_mod == 0
    tmp_mod %= popcount(tmp_mod)
    count += 1
  end
  ans.unshift(count)
end
ans.each { |a| puts a }