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: Time Limit Exceeded)

import math


N = int(input())
X_bin = input()

X_int = int(X_bin, 2)  # bin => int


def calc_bin_cnt(next_int):
    if next_int == 0:
        return 0
    '''
    next_bin = bin(next_int)[2:]
    # popcount
    cnt = 0
    for i in next_bin:
        if i == '1':
            cnt += 1
    '''
    cnt = 0
    n = int(math.log2(next_int))
    for i in range(n, -1, -1):
        if next_int >= 2**i:
            next_int -= 2**i
            cnt += 1
    return cnt


def calc_ans(next_int, cnt=None):
    if next_int == 0:
        return 0
    if cnt is None:
        cnt = calc_bin_cnt(next_int)
    next_int = next_int % cnt
    if next_int > 0:
        ans = calc_ans(next_int) + 1
    else:
        ans = 1
    return ans


X_cnt = calc_bin_cnt(X_int)
for i in range(N):
    if X_bin[i] == '1':
        print(calc_ans(X_int - (2**(N-i-1)), X_cnt-1))
    else:
        print(calc_ans(X_int + (2**(N-i-1)), X_cnt+1))