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: Wrong Answer)

import qualified Data.ByteString.Char8 as BS
import Data.Char
import Data.Bits
import Data.List

main = do
  li <- BS.getLine
  let Just (n,_) = BS.readInt li
  xs <- BS.getLine
  mapM_ print $ compute n xs

-- xsの2進数をbで割った余り
stepBS :: BS.ByteString -> Int -> Int
stepBS xs b = BS.foldl' (\acc c -> mod (acc+acc+digitToInt c) b) 0 xs

compute :: Int -> BS.ByteString -> [Int]
compute n xs = zipWith3 g (BS.unpack xs) rems0 rems1
  where
    (pc, pc0, pc1) = (BS.count '1' xs, succ pc, pred pc)
    rems0 = rems n pc0
    rems1 = rems n pc1
    stepBS0 = stepBS xs pc0
    stepBS1 = stepBS xs pc1
    g '0' rem0 _ = h pc0 stepBS0 rem0
    g '1' _ rem1 = h pc1 stepBS1 rem1
    h 0 _ _ = 0
    h pc stepbs rem = f $ mod (stepbs+rem) pc

--rems n 0 = replicate n 0 -- ??
rems n 1 = replicate n 0
rems n b = reverse $ take n $ iterate (\r -> (r*2) `mod` b) 1

step x = x `mod` popCount x

f = succ . length . takeWhile (0 /=) . iterate step
{-
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。
をちゃんと実装してみよう。
TLE
毎回 foldl (acc+acc+b `mod` pc1) するのは重いか。
(p,q) = divMod a b とは a = pb + q すると
divMod 2a b = (2p+x,y) where (x,y) = divMod 2q b
なぜなら 2q = bx+y, 2a = b(2p+x) + y = 2pb + bx + y = 2pb + 2q = 2(pb + q)
なので、xsを初めにsucc pcとpred pcで割った余りを求めておいて、
あとbit iの分は1から始めて2倍しては余りを求めるをi-1回繰り返した結果を求め、
さらに両者を足して余りをとればよい。
-}