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)

def cal_nums(number, num_one=-1):
    cp = number
    cnt = 0
    pos = 0
    if num_one == -1:
        while(1):
            if (cp >> pos) == 0:
                break
            cnt += ((cp>>pos)&1)
            pos += 1
    else:
        cnt = num_one
    
    # cnt_cp = cnt
    # mul = 2
    # while(cnt < number):
    #     cnt *= mul
    #     mul = mul << 1

    # return (number - cnt // mul) % cnt_cp
    return number % cnt

def func(number, num_one):
    cnt = 0
    tmp = number
    while(1):
        cnt += 1
        if cnt == 1:
            tmp = cal_nums(tmp, num_one)
        else:
             tmp = cal_nums(tmp)
        if tmp == 0:
            break
    return cnt

n = int(input())
string = input()
length = len(string)
tmp = 0
for i in range(length):
    tmp *= 2
    if string[i] == '1':
        tmp += 1

cnt_0 = 0
cnt_1 = 0

for idx in range(length):
    if string[idx] == '0':
        cnt_0 += 1
    else:
        cnt_1 += 1

for idx in range(length):
    if string[idx] == '0':
        print(func(tmp+(1<<(length-idx-1)), cnt_1+1))
        # print(tmp+(1<<(length-idx-1)))
    else:
        print(func(tmp-(1<<(length-idx-1)), cnt_1-1))
        # print(tmp-(1<<(length-idx-1)))