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: Accepted)

N = gets.to_i
X = gets.chomp.chars

def cnt(n, d)
  return d if n == 0
  wk = (0...n.bit_length).count{|i| n[i] == 1}
  cnt(n % wk, d+1)
end

c = X.count('1')
h, l = c+1, c-1
l = 1 if l == 0
t = Array.new(2).map{Array.new(N)}
t[0][N-1] = 1 % h
t[1][N-1] = 1 % l
m_h = (X[N-1] == '1' ? 1 : 0)
m_l = (X[N-1] == '1' ? 1 : 0)
(0..(N-2)).reverse_each{|n|
  t[0][n] = (t[0][n+1] << 1) % h
  t[1][n] = (t[1][n+1] << 1) % l
  if X[n] == '1' then
    m_h = (m_h + t[0][n]) % h
    m_l = (m_l + t[1][n]) % l
  end
}
m_l += l

(0...N).each{|n|
  if X[n] == '1' then
    if c == 1 then
      puts 0
    else
      puts cnt((m_l - t[1][n]) % l, 1)
    end
  else
    puts cnt((m_h + t[0][n]) % h, 1)
  end
}