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 Python (3.8.2) (judge verdict: Wrong Answer)

n=int(input())
x=input()
import sys
sys.setrecursionlimit(10**7)
if n==1:
  print(1 if x=='1' else 0)
  exit()
def pc(x):
  a=bin(x)[2:]
  return a.count('1')
def func(x):
  if x==0:
    return 0
  else:
    return func(x%pc(x))+1
pcx=x.count('1')
if pcx==1:
  print(0)
  for _ in range(n-2):
    print(1)
  print(2)
else:
  x0,x1=0,0
  for i in range(n):
    if x[i]=='1':
      x0+=pow(2,n-1-i,pcx-1)
      x0%=pcx-1
      x1+=pow(2,n-1-i,pcx+1)
      x1%=pcx+1
  for i in range(n):
    ans=0
    if x[i]=='1':
      xx=x0-pow(2,n-1-i,pcx-1)
      xx%=pcx-1
      print(func(xx)+1)
    else:
      xx=x1+pow(2,n-1-i,pcx+1)
      xx%=pcx+1
      print(func(xx)+1)