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
a = gets.split('').map(&:to_i)

li = []
for i in 0..n do
  cnt = 0
  p = Math.log(n,2).floor
  while p != -1
    if i>=2**p
      cnt += 1
      i -= p
    end
    p -= 1
  end
  li.push(cnt)
end

for i in 2..n do
  li[i] = li[i%li[i]]+1
end

###########################################
num = 0
for i in 0...a.size do
  num += a[i]
end

waru = [num-1,num+1]
amari = []
for i in waru do
  temp = []
  for j in 0...i do
    temp.push(2**j%i)
  end
  amari.push(temp)
end

sum = [0,0]
for i in 0...waru.size do
  for j in 0...a.size do
    sum[i] += a[a.size-1-j] * amari[i][j%waru[i]]
  end
end
sum[0] %= waru[0]
sum[1] %= waru[1]
##############################
ans = []
for i in 0...n do
  if a[a.size-1-i]==0
    ans.push((sum[1]+amari[1][i%waru[1]])%waru[1])
  else
    ans.push((sum[0]-amari[0][i%waru[0]])%waru[0])
  end
end

ans = ans.reverse
for i in 0...ans.size
  puts li[ans[i]]+1
end