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
x = gets.chomp.reverse

c = x.count(?1)

pow2_mod_up1 = Array.new(n)
pow2_mod_up1[0] = c + 1 < 2 ? 0 : 1
(n-1).times do |i|
    pow2_mod_up1[i+1] = pow2_mod_up1[i] << 1
    pow2_mod_up1[i+1] -= c + 1 if pow2_mod_up1[i+1] >= c + 1
end

pow2_mod_down1 = Array.new(n)
pow2_mod_down1[0] = c - 1 < 2 ? 0 : 1
(n-1).times do |i|
    pow2_mod_down1[i+1] = pow2_mod_down1[i] << 1
    pow2_mod_down1[i+1] -= c - 1 if c > 1 && pow2_mod_down1[i+1] >= c - 1
end

x_mod_up1 = x_mod_down1 = 0
n.times do |i|
    x_mod_up1 += pow2_mod_up1[i] * x[i].to_i
    x_mod_down1 += pow2_mod_down1[i] * x[i].to_i
end
x_mod_up1 %= c + 1
x_mod_down1 %= c - 1 if c > 1

n.times do |i|
    i = n - i - 1
    if x[i].to_i == 0
        a = (x_mod_up1 + pow2_mod_up1[i]) % (c + 1)
    else
        a = c <= 1 ? 0 : (x_mod_down1 + pow2_mod_down1[i]) % (c - 1)
    end

    ans = 1
    while a != 0
        a %= a.to_s(2).each_char.count(?1)
        ans += 1
    end
    p ans
end