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)

require 'set'; require 'prime'; require 'pp'
INF=Float::INFINITY; MOD=10**9+7
n = gets.chomp.to_i
xs = gets.chomp

# n = 200000
# xs = '11010110111100001010' * 10000

xi = xs.to_i(2)
init_count = xs.count('1')

xi_init = xi
xs_init = xs.dup

# p [xi_init, xs_init]

xi_init_m1 = xi_init % (init_count-1)
xi_init_p1 = xi_init % (init_count+1)

anss=[]
hiku_m1 = 1
hiku_p1 = 1
n.times do |i|
  if xs_init[n-i-1] == '1'
    if init_count == 1
      anss.push 0
      next
    end
    count = init_count - 1
    # xi = xi_init_m1 - mod_kai(n-i-1, count)
    xi = xi_init_m1 - hiku_m1
  else
    count = init_count + 1
    # xi = xi_init_p1 + mod_kai(n-i-1, count)
    xi = xi_init_p1 + hiku_p1
  end

  # p [i, [xi, xi.to_s(2)], count]
  xi = (xi + count) % count

  xi %= count #=
  xs = xi.to_s(2)

  ans = 1
  while xi > 0
    count = xs.count('1')
    xi %= count #=
    xs = xi.to_s(2)
    ans += 1
  end
  hiku_m1 = hiku_m1 * 2 % (init_count-1)
  hiku_p1 = hiku_p1 * 2 % (init_count+1)
  anss.push ans
end

anss.reverse_each do |e|
  puts e
end