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

def popcount(a)
  m1 = 0x55555555
  m2 = 0x33333333
  m4 = 0x0f0f0f0f
  a -= (a >> 1) & m1
  a = (a & m2) + ((a >> 2) & m2)
  a = (a + (a >> 4)) & m4
  a += a >> 8
  return (a + (a >> 16)) & 0x3f
end

hash = {}
max = 2 * (10 ** 5)
(1..max).each do |i|
  pc = popcount(i)
  tmp = i % pc
  if tmp == 0
    count = 1
  else
    count = hash[tmp] + 1
  end
  hash[i] = count
end

tmp1 = 0
tmp2 = 0
pc = popcount(x.to_i(2))
(0..n-1).each do |i|
  if x[i] == "1"
    tmp1 += 2.pow(n-i-1, pc + 1)
    tmp1 %= (pc + 1)
    if pc > 1
      tmp2 += 2.pow(n-i-1, pc - 1)
      tmp2 %= (pc - 1)
    end
  end
end

(0..n-1).each do |i|
  if x[i] == "0"
    tmp = tmp1
    tmp += 2.pow(n-i-1, pc + 1)
    tmp %= (pc + 1)
  else
    if pc > 1
      tmp = tmp2
      tmp -= 2.pow(n-i-1, pc - 1)
      tmp %= (pc - 1)
    else
      puts 0
      next
    end
  end
  puts hash[tmp].to_i + 1
end