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: Time Limit Exceeded)

N = gets.to_i
X = gets.chomp
# N = 2 * 10**5
# X = "1"*(N-1) + "0"
xcnt = X.count('1')

@fmap = {
    0 => 0,
    1 => 1
}

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

def f x
    return @fmap[x] if @fmap[x]
    pc = popcount(x)
    if pc == 0
        @fmap[x] = 0
        return 0
    end
    m = x % pc
    if @fmap[m]
        @fmap[x] = @fmap[m] + 1
    else
        @fmap[x] = f(m) + 1
    end
    @fmap[x]
end

diff = 1
intx = X.to_i(2)
xar = X.split('').reverse
pl_mod = intx % (xcnt + 1)
mi_mod = false
if xcnt > 1
    mi_mod = intx % (xcnt - 1)
end
results = xar.map do |i|
    mo = nil
    if i == "0"
        mo = (pl_mod + diff) % (xcnt + 1)
    elsif !mi_mod
        next 0
    else
        mo = (mi_mod - diff) % (xcnt - 1)
    end
    next f(mo) + 1
ensure
    diff *= 2
end

puts results.reverse.join("\n")