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 Fortran (GNU Fortran 9.2.1) (judge verdict: Runtime Error)

program anything_goes_to_zero
  implicit none
  integer :: n, i, m, k, cnt = 0, rem1to0 = 0, rem0to1 = 0, rem
  character(200000) :: x
  read(*,*) n
  read(*,*) x
  do i = 1, n
    if (x(i:i) == '1') cnt = cnt + 1
  end do
  do i = 1, n
    k = ichar(x(i:i)) - 48
    rem1to0 = mod(2 * rem1to0 + k, cnt - 1)
    rem0to1 = mod(2 * rem0to1 + k, cnt + 1)
  end do
  do i = 1, n
    if (x(i:i) == '1') then
      rem = modulo(rem1to0 - pow(2, n - i, cnt - 1), cnt - 1)
    else
      rem = mod(rem0to1 + pow(2, n - i, cnt + 1), cnt + 1)
    end if
    m = 1
    do while (rem > 0)
      k = popcnt(rem)
      rem = mod(rem, k)
      m = m + 1
    end do
    write(*,'(i0)') m
  end do
contains
  pure elemental integer function pow(a, n, modu) result(res)
    integer, intent(in) :: a, n, modu
    integer :: b, m
    res = 1
    b = a
    m = n
    do while (m > 0)
      if (btest(m, 0)) res = mod(res * b, modu)
      b = mod(b * b, modu)
      m = rshift(m, 1)
    end do
  end
end program anything_goes_to_zero