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)

io = DATA.eof? ? STDIN : DATA
n=io.gets.to_i
xs=io.gets.chomp
x=xs.to_i(2)
def popcount(x)
  b = 0
  while x > 0
    x &= x - 1
    b += 1
  end
  b
end
def f(n)
  cnt=1
  while (n>0)
    n=n%popcount(n)
    cnt+=1
  end
  cnt
end
pcx=popcount(x)
pc0,pc1= pcx+1,pcx-1
xmod0=x%pc0
xmod1=x%pc1 if pc1>0
n.times do |i|
  if (xs[i])=="0"
    pci,xmodi,powi = [pc0,xmod0,(2.pow(n-1-i,pc0))]
  elsif pc1>0
    pci,xmodi,powi = [pc1,xmod1,-(2.pow(n-1-i,pc1))]
  else
    pci=pc1
  end
  if pci==0
    puts 0
  else
    pc=(xmodi+powi)%pci
    puts f(pc)
  end
end
__END__