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


def popcount(v)
    v = v - ((v >> 1) & 0x55555555)
    v = (v & 0x33333333) + ((v >> 2) & 0x33333333)
    ((v + (v >> 4) & 0xF0F0F0F) * 0x1010101) >> 24
end

max = 2*10**5
arr = Array.new(max+1,0)
1.upto(max) do |i|
    arr[i] = 1 + arr[i%popcount(i)]
end

pcount_base = S.count('1')
xa = X % (pcount_base+1)
xb = X % [(pcount_base-1),1].max
1.upto(N) do |i|
    pcount = pcount_base + (S[i-1] == '1' ? -1 : 1)
    if pcount == 0
        puts 0
    elsif S[i-1] == '1'
        x = xb - ((2**(N-i)) % pcount)
        puts 1+arr[x%pcount]
    else
        x = xa + ((2**(N-i)) % pcount)
        puts 1+arr[x%pcount]
    end
end