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 Haskell (GHC 8.8.3) (judge verdict: Time Limit Exceeded)

main = do
    getLine
    str <- getLine
    let ls = map b2i $ okikaeList str
    mapM_ print $ map func ls
    
popcount 0 = 0
popcount 1 = 1
popcount x
    | mod x 2 == 0  = popcount $ div x 2
    | otherwise     = 1 + (popcount (div x 2))
    
func n
    | n == 0        = 0
    | pop == 0      = 1
    | otherwise     = 1 + func m
    where pop = popcount n
          m   = mod n pop
          
okikae '0' = '1'
okikae '1' = '0'
          
okikaeN str n = (take n str) ++ [okikae (str!!n)] ++ (drop (n+1) str)

okikaeList str = zipWith okikaeN (replicate (length str) str) [0,1..]

b2i xs = foldl (\x y -> 2 * x + y) 0 $ map (\x -> read [x]) xs