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
def func(z,zzzz)
  return (z.to_i(2))%zzzz
end
aa = Array.new(n+1)
m = x.count("1")

f = [["0",0],["1",1]]
while f.size > 0
  if f[0][1] != 0 && f[0][0].to_i(2) <= n
    b = func(f[0][0],f[0][1])
    if b == 0
      aa[f[0][0].to_i(2)-1] = 1
    else
      aa[f[0][0].to_i(2)-1] = 1+aa[b-1]
    end
  end
  f << ["0"+f[0][0],f[0][1]] if ("0"+f[0][0]).to_i(2) <= n && ("0"+f[0][0]).size < 19
  f << ["1"+f[0][0],f[0][1]+1] if ("1"+f[0][0]).to_i(2) <= n
  f.shift
end

ans = []

for j in 0..(n-1)
  if x[j] == "0"
    x[j] = "1"
    b = x.to_i(2) % (m+1)
    if b == 0
      ans << 1
    else
      ans << 1+aa[b-1]
    end
    x[j] = "0"
  else
    x[j] = "0"
    b = x.to_i(2) % (m-1)
    if b == 0
      ans << 1
    else
      ans << 1+aa[b-1]
    end
    x[j] = "1"
  end
end
puts ans