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 PyPy3 (7.3.0) (judge verdict: Runtime Error)

import sys
input=lambda: sys.stdin.readline().rstrip()
n=int(input())
X=input()
def popcount(x):
  x=x-((x>>1)&0x5555555555555555)
  x=(x&0x3333333333333333)+((x>>2)&0x3333333333333333)

  x=(x+(x>>4))&0x0f0f0f0f0f0f0f0f
  x=x+(x>>8)
  x=x+(x>>16)
  x=x+(x>>32)
  return x&0x0000007f

F=[0]*(n+1)
F[1]=1
F[2]=1
for i in range(3,n+1):
  F[i]=1+F[i%popcount(i)]

ct=X.count("1")
if ct==0:
  for _ in range(n):
    print(1)
else:
  a,b=0,0
  for i in range(n):
    if X[n-1-i]=="1":
      a=(a+pow(2,i,ct-1))%(ct-1)
      b=(b+pow(2,i,ct+1))%(ct+1)
  for i in range(n):
    if X[i]=="0":
      print(F[(b+pow(2,n-1-i,ct+1))%(ct+1)]+1)
    else:
      print(F[(a-pow(2,n-1-i,ct-1))%(ct-1)]+1)