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: Time Limit Exceeded)

(defun f (x)
  (let ((acc 0))
    (loop :while (< 0 x)
          :do (setf x (mod x (logcount x)))
          :do (incf acc))
    acc))

(let* ((n (read))
       (s (read-line))
       (x (parse-integer s :radix 2))
       (p (logcount x))
       (2+ (make-array (list n)))
       (2- (make-array (list n)))
       (y+ 0)
       (y- 0))
  (setf (aref 2+ 0) (mod 1 (1+ p)))
  (loop :for i :from 1 :to (1- n)
        :do (setf (aref 2+ i) (mod (* (aref 2+ (1- i)) 2) (1+ p))))
  (setf y+ (mod x (1+ p)))
  (when (/= p 1)
    (setf (aref 2- 0) (mod 1 (1- p)))
    (loop :for i :from 1 :to (1- n)
          :do (setf (aref 2- i) (mod (* (aref 2- (1- i)) 2) (1- p))))
    (setf y- (mod x (1- p))))
  (format t
          "~{~A~%~}" 
          (loop :for i :from 0 :to (1- n)
                :for j :downfrom (1- n) :to 0
                :collect (cond ((and (char= (aref s i) #\1) (= p 1))
                                0)
                               ((char= (aref s i) #\1)
                                (1+ (f (mod (- y- (aref 2- j)) (1- p)))))
                               (t
                                (1+ (f (mod (+ y+ (aref 2+ j)) (1+ p)))))))))