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)

# frozen_string_literal: true

N = gets.chomp.to_i
X = gets.chomp.to_i(2)

memo = {}

def popcount(x)
  # 拝借したポップカウント
  x -= ((x >> 1) & 6_148_914_691_236_517_205)

  x = (x & 3_689_348_814_741_910_323) + ((x >> 2) & 3_689_348_814_741_910_323)

  x = (x + (x >> 4)) & 1_085_102_592_571_150_095
  x += (x >> 8)
  x += (x >> 16)
  x += (x >> 32)
  x & 127
end

def f(n)
  count = 0
  until n.zero?
    count += 1
    n = n % popcount(n)
  end
  count
end

def popswitch(x, i)
  num = 1 << (i - 1)
  x ^ num
end

(1..N).reverse_each do |n|
  num = popswitch(X, n)
  memo[num] ||= f(num)
  puts memo[num]
end