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: Wrong Answer)

def solve():
    N = int(input())
    X = input()
    X_10 = int(X,2)

    r = X.count("1")
    print(r)
    r0 = r-1
    r1 = r+1
    ans = []
    
    s0 = X_10 % r0 if r != 1 else None
    s1 = X_10 % r1 

    for i in range(N):
        if X[i] == '0':
            ans.append(f((s1 + pow(2,N-1-i,r1)) % r1))
        else:
            if r0 == 0:
                ans.append(0)
            else:
                ans.append(f((s0 - pow(2,N-1-i,r0)) % r0))
    
    print(*ans, sep='\n')

def f(N):
    cnt = 1
    while N != 0:
        N = N % popcount(N)
        cnt += 1
    
    return cnt

def popcount(N):
    b = bin(N)
    cnt = 0
    for i in range(2,len(b)):
        if b[i] == '1':
            cnt += 1

    return cnt

if __name__ == '__main__':
    solve()