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

N = int(input())
X = input()
px = X.count('1')

def xmod(m):
    res = []
    mod = 1
    for s in reversed(X):
        res.append(mod)
        mod = (mod * 2) % m
    
    return list(reversed(res))

plus = xmod(px+1)
minus = xmod(px-1)
sp = sum(plus)
sm = sum(minus)

for i in range(N):
    if X[i] == '0':
        sum_ = (sp - plus[i]) % (px+1)
        if sum_ == 0:
            print(0)
        else:
            cnt = 0
            while sum_ == 0:
                cnt += 1
                sum_ = sum_ % popcpunt(sum_)
            print(cnt)
    else:
        sum_ = (sm - minus[i]) % (px-1)
        if sum_ == 0:
            print(0)
        else:
            cnt = 0
            while sum_ == 0:
                cnt += 1
                sum_ = sum_ % popcpunt(sum_)
            print(cnt)
      
def popcount(x):
    x = x - ((x >> 1) & 0x5555555555555555)

    x = (x & 0x3333333333333333) + ((x >> 2) & 0x3333333333333333)

    x = (x + (x >> 4)) & 0x0f0f0f0f0f0f0f0f # 8bitごと
    x = x + (x >> 8) # 16bitごと
    x = x + (x >> 16) # 32bitごと
    x = x + (x >> 32) # 64bitごと = 全部の合計
    return x & 0x0000007f