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: Runtime Error)

N = gets.to_i
X = gets.chomp

pop_cnt = X.count('1')

p0 = pop_cnt + 1
p1 = pop_cnt - 1
bases = [X.to_i(2) % p0, X.to_i(2) % p1]

$cache = Hash.new
$cache[0] = 0
 
class Integer
  def popcount
    to_s(2).count('1')
  end
end
 
def f(x)
  return $cache[x] if $cache[x]

  $cache[x] = f(x % x.popcount) + 1
end

v0 = 1 % p0
v1 = 1 % p1

ans = []

N.times do |i|
  if X[N - i - 1] == '0'
    r = (bases[0] + v0) % p0
  else
    r = (bases[1] - v1) % p1
  end

  if r == 0
    ans << 1
  else
    ans << f(r) + 1
  end

  v0 = (2 * v0) % p0
  v1 = (2 * v1) % p1
end

puts ans.reverse