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: Wrong Answer)

n = int(input())
x = input()

x0 = int('0b'+x,0)

ceil = 2*10**5+2
ppc = [0]*(ceil)

def popcount(n):    
  if n == 0:
    return 0
  if ppc[n] != 0:
    return ppc[n]
    
  b = bin(n).count('1')  
  tmp = n%b
  ppc[n] = popcount(tmp) + 1
  return ppc[n]

c1 = bin(x0).count('1')  
p = 1
ans = [0]*n

if c1-1 == 0:
  x0m1 = 0
else:  
  x0m1 = x0 % (c1 - 1)
x0m0 = x0 % (c1 + 1)

for i in range(n): 
  
  if x[n-i-1] == '1':    
    if x0m1 == 0:
      ans[i] = 0
      continue
      
    tmp = (x0m1 - p) % (c1 - 1)
  
    if c1-1 == 0:
      tmp = 0
    else:
      tmp2 = tmp % (c1 - 1)
      
  else:    
    tmp = (x0m0 + p) % (c1 + 1)
    tmp2 = tmp % (c1 + 1)
    
  if tmp > ceil:      
    ans[i]=popcount(tmp2)+2
  else:
    ans[i]=popcount(tmp)+1
  p *= 2
  
for i in range(n):
  print(ans[-1-i])