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: Runtime Error)

import numpy as np
import math
import scipy as sp
import queue
import collections

MOD = 10 ** 9 + 7
INF = 10 ** 9

def main(kwargs):
    n = kwargs["n"]
    x = kwargs["x"]
    F = [0 for i in range(int("1" * n, 2))]

    def f(x):
        if F[x - 1] != 0:
            return F[x - 1]
        else:
            new_x = x % popcount(x)

            if new_x == 0:
                return 1
            else:
                F[x - 1] = f(new_x) + 1
                return F[x - 1]

    for i in range(int("1" * n, 2)):
        F[i] = f(i + 1)

    X = [x for i in range(n)]

    for i in range(1, n):
        if X[i] == "0":
            X[i] = "{}1{}".format(X[i][:i], X[i][i + 1:])
        else:
            X[i] = "{}0{}".format(X[i][:i], X[i][i + 1:])

    res = [F[int(x, 2) - 1]for x in X]
    res = "\n".join(res)
    return res

def popcount(x):
    s = bin(x)
    return s.count("1")

if __name__ == "__main__":
    kwargs = {}

    cin = input().split()
    kwargs["n"] = [int(i) for i in cin][0]
    cin = input().split()
    kwargs["x"] = cin[0]

    cout = main(kwargs)
    print(cout)