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: Accepted)

def popcount(x)
    p = 0
    while x > 0
        x &= x - 1
        p += 1
    end
    return p
end

N = gets.to_i
X = gets.to_i(2)
ans = Array.new(N,0)

p0 = 0
N.times do |i|
    p0 += 1 if X[i] == 1
end


x0_mod_pp1 = X % (p0+1)
x0_mod_pm1 = X % (p0-1) if p0 != 1

N.times do |i|
    next if X ^ (1 << i) == 0
    pi = p0
    r = 0
    if X[i] == 1
        pi -= 1
        r = (x0_mod_pm1 - 2.pow(i, pi) + pi) % pi
        ans[i] += 1
    else
        pi += 1
        r = (x0_mod_pp1 + 2.pow(i, pi)) % pi
        ans[i] += 1
    end
    
    # r = xi
    # p r
    while r != 0 do
        pi = popcount(r)
        r = r % pi
        ans[i] += 1
    end
end
puts ans.reverse