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: Time Limit Exceeded)

N=int(input())
S=input()
X=int(S,2)

#Sのビット数
bitcount=0
for i in range(N):
  if S[i]=='1':
    bitcount+=1

def f(n):
  c=0
  x=1
  while x<=n:
    if n&x>0:
      c+=1
    x*=2
  return n%c

ls=[0 for i in range(bitcount+2)]  
ls2=[0 for i in range(bitcount+2)]
for i in range(1,bitcount+2):
  ls[i]=f(i)
  ls2[i]=ls2[ls[i]]+1

for i in range(N):
  if S[i]=='0':
    print(ls2[(X^(2**(N-1-i)))%(bitcount+1)]+1)
  else:
    if (X^(2**(N-1-i)))==0:
      print(0)
    else:
      print(ls2[(X^(2**(N-1-i)))%(bitcount-1)]+1)