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 Python (3.8.2) (judge verdict: Accepted)

def main():
    N = int(input())
    X = input()
    XINT = int(X,2)
    ANS = [0]*N
    TEMP = [0]*N
    CHAIND = [0]*N

    cnt = count_ones_by_bitmask(XINT)
    
    if cnt == 0:
        for i in range(N):
            print(1)
            return
    if cnt ==1:
        for i in  range(N):
            if X[i] == '1':
                print(0) #全てが0になる
            else:
                if i==N-1:
                    print((int(X[N-1])+1)%2+1) # 2の剰余の反転
                else:
                    print((int(X[N-1]))%2+1) # 2の剰余
        return

    p = XINT % (cnt+1)
    m = XINT % (cnt-1)
    pmod = 1
    mmod = 1
    for i in range(N):
        ANS[i] = 1
        if (X[N-i-1] == '0'):
            TEMP[i] = (p + pmod) % (cnt+1)
        else:
            TEMP[i] = (m - mmod) % (cnt-1)
        pmod = (pmod*2) % (cnt+1)
        mmod = (mmod*2) % (cnt-1)


    CHA = list(set(TEMP))

    for i in range(len(CHA)):
        an = 0
        cha = CHA[i]
        cnt = count_ones_by_bitmask(CHA[i])
        while cnt>0:
            an += 1
            cha = cha%cnt
            cnt = count_ones_by_bitmask(cha)
        CHAIND[CHA[i]] = an

    for i in reversed(range(N)):
        print(ANS[i]+CHAIND[TEMP[i]])


def count_ones_by_bitmask(num):
    count = 0
    while num:
        count += num & 1
        num >>= 1
    return count

if __name__ == '__main__':
    main()