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.map &:to_i
b=x.count(1)
if b==0
  n.times{ puts 1 }
  exit
end
h=[0]
f=->x{
  t=h[x] and return t
  pc=x.to_s(2).chars.count(?1)
  h[x]=f[x%pc]+1
}
#p [n,x,b]
rp=rm=0
rps,rms=[0]*n,[0]*n
tm=tp=1
if b>2
  (n-1).downto(0){|i|
    rps[i]=tp
    rms[i]=tm
    if x[i]==1
      rp=(rp+tp)%(b+1)
      rm=(rm+tm)%(b-1)
    end
    tp=tp*2%(b+1)
    tm=tm*2%(b-1)
  }
else
  (n-1).downto(0){|i|
    rps[i]=tp
    if x[i]==1
      rp=(rp+tp)%(b+1)
    end
    tp=tp*2%(b+1)
  }
end
#p [rp,rm,rps,rms]
n.times{|i|
  if x[i]==0
    puts 1+f[(rp+rps[i])%(b+1)]
  elsif b==1
    puts 0
  else
    puts 1+f[(rm-rms[i])%(b-1)]
  end
}