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,x = gets.to_i,gets
c,x = x.scan(?1).count, x.to_i(2)
R = [1,-1].map { m = c+_1; x%m if m>1 }
M = {0=>1}
X = (N-1).downto(0).map {|n|
  b = x[n]
  r,m = R[b],c+1-b-b
  if m < 2
    m - 2
  elsif b > 0
    (0 > r-=2.pow(n,m)) ? r+m : r
  else
    (c < r+=2.pow(n,m)) ? r-m : r
  end
}
I = [*0...N].sort_by! {X[_1]}
S = []
I.each {|i|
  next  if 1 > x=X[i]
  until c=M[x]
    S << x
    b,c = x+x,0
    c+=b[0] while 0 < (b>>=1)
    x %= c
  end
  M[x] = c+=1  while x=S.pop
}
X.each {|x| p x<0 ? x+2 : M[x]}