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 Ruby (2.7.1) (judge verdict: Wrong Answer)

n = gets.chomp.to_i
x = gets.chomp
xi = x.to_i(2)
xc = x.count('1')

def exec(n)
  c = 0
  tmp = n
  while tmp != 0
    c += 1 if tmp % 2 == 1
    tmp /= 2
  end

  n % c
end

memo = {}
1.upto(n) do |n|
  tmp = n
  ans = 0
  while tmp != 0
    ans += 1
    tmp = exec(tmp)
  end

  memo[n] = ans
end
memo[0] = 0

solver = ->(i) do
  del = 2 ** (n-i-1)
  zero_p = x[i] == '0'
  num = zero_p ? xi + del : xi - del

  if num < 20
    memo[num]
  else
    if zero_p
      c = xc + 1
    else
      c = xc - 1
    end
    memo[num % c] + 1
  end
end

n.times do |i|
  puts solver.call(i)
end