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)

#!/usr/bin/env python3

def main():
    N = int(input())
    X = [None] * N
    l = list(input())
    Xc = 0
    for i in range(N):
        X[i] = int(l[i])
        Xc += X[i]
    if Xc == 0:
        print(0)
        exit()
    elif Xc == 1:
        print(1)
        exit()

    ans1 = 0
    ans2 = 0
    for j in range(N):
        if X[j]==1:
            ans1 += pow(2,N-1-j,Xc+1)
            ans1 %= Xc+1
            ans2 += pow(2,N-1-j,Xc-1)
            ans2 %= Xc-1

    for i in range(N):
        if X[i]==0:# 0->1
            ans = (ans1 + pow(2,N-1-i)) % (Xc+1)
        else:# 1-> 0
            ans = (ans2 - pow(2,N-1-i)) % (Xc-1)
        count = 1
        while ans > 0:
            tmp = str(bin(ans)).count('1')
            ans %= tmp
            count += 1
        print(count)

if __name__ == "__main__":
    main()