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: Time Limit Exceeded)

#require 'prime'
#require 'openssl'
MOD=10**9+7;  cnt=0; sum=0; prev=nil; can=true; h=Hash.new(0)
def gs() gets.chomp end
def gi() gets.chomp.to_i end
def gsmi() gets.chomp.split.map(&:to_i) end

def ar2(s1,s2,de=nil) Array.new(s1){Array.new(s2,de)} end
def ar3(s1,s2,s3,de=nil) Array.new(s1){Array.new(s2){Array.new(s3,de)}} end

def desc(ar) ar.sort!{|x,y| y<=>x} end
def min(a,b) a<=b ? a : b end
def max(a,b) a>=b ? a : b end
def sum(ar) ar.inject(:+).to_i end
def ceil(a,b) (a.to_f/b).ceil end

def C(a,b) b==0||a==b ? 1 : (b=a-b if a/2<b;(a-b+1..a).inject(:*)/(1..b).inject(:*)) end
def rui(ar) s=[0]; ar.size.times{|i| s<<s[-1]+ar[i]}; s end
def rev_rui(ar) s=[0]; ar.size.times.reverse_each{|i| s<<s[-1]+ar[i]}; s end

def puts_yesno(b) puts(b ? 'Yes' : 'No') end
def putsend(s) puts s; exit end
def pu(h) begin; require './../../abc/nemi_lib/debug.rb'; puts_val(h); rescue LoadError; end; end
########### ( ˘ω˘ )ｽﾔｧ… ###########
n=gi
x=gs
Memo={}
def execute(n,cnt=1,i_bit=nil)
  return 0 if n==0
  pc=
    if Memo[n]
      Memo[n]
    else
      if i_bit
        i_bit
      else
        n.to_s(2).count("1")
      end
    end
  Memo[n]=pc
  mod=n%pc
  return cnt if mod==0
  execute(mod,cnt+1)
end

i_bit=x.count("1")
ans=[]
n.times do |i|
  xi=Marshal.load(Marshal.dump x)
  xi[i]= (xi[i]=="1" ? "0" : "1")
  num = xi.to_i 2
  pu num:num
  ans<<execute(num,1,(xi[i]=="1" ? i_bit+1 : i_bit-1))
end
puts ans