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 PyPy3 (7.3.0) (judge verdict: Accepted)

n = int(input())
x = input()
mod_num1 = [0 for i in range(n+1)]
mod_num2 = [0 for i in range(n+1)]
mod_num3 = [0 for i in range(n+1)]
num2 = x.count('1')
if num2 == 0:
    for i in range(n):
        print(1)
    exit()
elif num2 == 1:
    index = x.index('1')
    for i in range(n):
        if x[i] == '1':
            print(0)
        else:
            if index == n-1 or i == n-1:
                print(2)
            else:
                print(1)
    exit()
num1 = num2 - 1
num3 = num2 + 1
for i in range(n):
    if x[i] == '1':
        mod_num1[i] = pow(2, n-i-1 ,num1)
        mod_num2[i] = pow(2, n-i-1, num2)
        mod_num3[i] = pow(2, n-i-1, num3)
def popcount(num):
    res = 0
    while num != 0:
        res += 1
        num = num % "{:b}".format(num).count('1')
    return res
s_1 = sum(mod_num1)
s_2 = sum(mod_num2)
s_3 = sum(mod_num3)
for i in range(n):
    if x[i] == '1':
        print(popcount((s_1 - pow(2, n-i-1, num1)) % num1) + 1)
    else:
        print(popcount((s_3 + pow(2, n-i-1, num3)) % num3) + 1)