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 OCaml (4.10.0) (judge verdict: Runtime Error)

let popcnt x =
  let x = (x land 0x5555555555555555) + ((x lsr 1) land 0x5555555555555555) in
  let x = (x land 0x3333333333333333) + ((x lsr 2) land 0x3333333333333333) in
  let x = (x land 0x0f0f0f0f0f0f0f0f) + ((x lsr 4) land 0x0f0f0f0f0f0f0f0f) in
  let x = (x land 0x00ff00ff00ff00ff) + ((x lsr 8) land 0x00ff00ff00ff00ff) in
  let x = (x land 0x0000ffff0000ffff) + ((x lsr 16) land 0x0000ffff0000ffff) in
  (x land 0x00000000ffffffff) + ((x lsr 32) land 0x00000000ffffffff)

let rec power ( * ) e m n =
  if n <= 0 then e
  else power ( * ) (if n land 1 = 0 then e else m * e) (m * m) (n lsr 1)

let () = Scanf.scanf "%d\n%s" @@ fun n x ->
  let x = Array.init n @@ String.get x in
  let dp = Array.make (n + 2) 0 in
  for i = 1 to n + 1 do
    dp.(i) <- 1 + dp.(i mod popcnt i)
  done;
  let pop = Array.fold_right (function '0' -> Fun.id | '1' -> succ) x 0 in
  let m0 = Array.fold_left (fun acc c -> (2 * acc + if c = '0' then 0 else 1) mod (pop + 1)) 0 x in
  let m1 = Array.fold_left (fun acc c -> (2 * acc + if c = '0' then 0 else 1) mod (pop - 1)) 0 x in
  Fun.flip Array.iteri x @@ fun i c ->
    let i = n - i - 1 in
    Printf.printf "%d\n" @@
    succ @@
    Array.get dp @@
    if c = '0'
    then (m0 + power (fun x y -> (x * y) mod (pop + 1)) 1 2 i) mod (pop + 1)
    else (m1 + pop - 1 - power (fun x y -> (x * y) mod (pop - 1)) 1 2 i) mod (pop - 1)