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 Common Lisp (SBCL 2.0.3) (judge verdict: Wrong Answer)

(defvar *dp*
  (make-hash-table :size (expt 10 8)))

(defun pop-count (x &optional (acc 0) (l nil))
  (let ((z (loop :for c :across (format nil "~B" x) :if (char= c #\1) :sum 1)))
    (cond ((= x 0)
           (loop :for y :in l
                 :for i :from 1
                 :do (setf (gethash y *dp*) (- acc i))) 
           acc)
          ((gethash x *dp*)
           (+ (gethash x *dp*) acc))
          (t
           (pop-count (mod x z) (1+ acc) (cons (mod x z) l))))))

(let* ((n (read))
       (_ (read-line))
       (x (parse-integer _ :radix 2)))
  (loop :for i :downfrom (1- n) :to 0
        :do (let* ((y (if (> (logand x (expt 2 i)) 0)
                          (- x (expt 2 i))
                          (+ x (expt 2 i))))
                   (ans (pop-count y)))
              (setf (gethash i *dp*) ans)
              (format t "~A~%" ans))))