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)

n = int(input())
x = input()
cnt1 = x.count('1')
x10 = int(x,2)

if cnt1 > 1:
  x10mod_m = x10 % (cnt1-1)
  x10mod_p = x10 % (cnt1+1)
else:
  x10mod_m = 0
  x10mod_p = 0

ans = []
for i in range(n):
  add_val = add_val >> 1
  if x[i] == '1':
    if cnt1 == 1:
      print(0)
      continue
    cntx = cnt1-1
    mod = ( x10mod_m - pow(1,n-1-i,cntx) ) % cntx
  else:
    cntx = cnt1+1
    mod = ( x10mod_p + pow(1,n-1-i,cntx) ) % cntx
    
  cnt = 1
  mod = format(mod, 'b')
  
  while True:
    if mod == '0':
      break
    mod = format( int(mod,2) % mod.count('1'), 'b')
    cnt += 1
  print(cnt)