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: Accepted)

import Data.Bits
import Data.Char
import Data.List

main = do
  li <- getLine
  let n = read li
  xs <- getLine
  mapM_ print $ compute n xs

compute :: Int -> String -> [Int]
compute n xs = [ f (x `xor` bit i) | i <- [n-1,n-2..0] ]
  where
    x :: Integer
    x = foldl' (\acc b -> acc+acc+b) 0 $ map (fromIntegral . digitToInt) xs

step :: Integer -> Integer
step x = x `mod` fromIntegral (popCount x)

f :: Integer -> Int
f x = length $ takeWhile (0 /=) $ iterate step x
{-
最大2万桁の2進数を扱うのは、Integerでやればできるかもしらんがどうかという。
最初のpopCountはたかだか2万なので、あまりもそれ未満。だから初回さえ乗り切ればあとは普通の整数で間に合う。
けどそれも面倒なので、IntegerのData.Bitsでゴリ押ししてみよう。
-}