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)

n = gets.to_i
x = gets.chomp.chars
pc = x.count("1")

ur = []
lr = []
ut = 1
lt = -1
n.times do
  ut %= pc + 1
  ur.unshift(ut)
  ut <<= 1
  if pc - 1 > 0
    lt %= pc - 1
    lr.unshift(lt)
    lt <<= 1
  end
end

rl = x[0].to_i
ru = x[0].to_i
1.step(n - 1) do |i|
  t = x[i]
  if pc - 1 > 0
    rl <<= 1
    rl += t.to_i
    rl %= pc - 1
  end
  ru <<= 1
  ru += t.to_i
  ru %= pc + 1
end

n.times do |i|
  res = 1
  if x[i] == "0"
    t = (ur[i] + ru) % (pc + 1)
  elsif pc - 1 == 0
    puts "0"
  else
    t = (lr[i] + rl) % (pc - 1)
  end
    
  while t > 0
    t %= ("%b" % [t]).count("1")
    res += 1
  end
  puts res
end