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)

def bc(x):
    return bin(x).count("1")

def f(n):
    res = 0
    for i in range(n):
        if n == 0:
            return res
        else:
            res += 1
            n = n % bc(n)
    return res

N,X = open(0).read().split()
N = int(N)
pc = X.count('1')
na = 0
nb = 0
if N == 1:
    if X == '1':
        print(0)
    else:
        print(1)
else:
    if pc == 0:
        for i in range(N):
            print(1)
    elif pc == N:
        for i in range(1,N+1):
            na = (na+pow(2,N-i,N-1)) % (N-1)
        for i in range(1,N+1):
            temp = na
            temp = (temp-pow(2,N-i,N-1)) % (N-1)
            print(f(temp)+1)
    elif pc == 1:
        ind = X.index('1')
        na = pow(2,N-ind-1,2)
        for i in range(1,N+1):
            if i != ind + 1:
                temp = na
                temp = (temp+pow(2,N-i,2)) % 2
                print(f(temp)+1)
            else:
                print(0)
    else:
        for i in range(1,N+1):
            if X[i-1] == '1':
                na = (na+pow(2,N-i,pc-1)) % (pc-1)
        for i in range(1,N+1):
            if X[i-1] == '1':
                nb = (nb+pow(2,N-i,pc+1)) % (pc+1)
        for i in range(1,N+1):
            if X[i-1] == '1':
                temp = na
                temp = (temp-pow(2,N-i,pc-1)) % (pc-1)
                print(f(temp)+1)
            else:
                temp = nb
                temp = (temp+pow(2,N-i,pc+1)) % (pc+1)
                print(f(temp)+1)