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: Accepted)

import sys
sys.setrecursionlimit(200000)
n=int(input())
x=input()

tab=[0]+[i%bin(i).count('1') for i in range(1,200001)]

xcnt=x.count("1")
staxp=0
staxm=0
pl=xcnt+1
mi=xcnt-1
if mi==0:
    mi=1
for i in range(n):
    if x[n-i-1]=="1":
        staxp=(staxp+pow(2,i,pl))%pl
        staxm=(staxm+pow(2,i,mi))%mi
    
def calc(x,now):
    if x==0:
        return now
    else:
        return calc(tab[x],now+1)

for i in range(n):
    if x[i]=="1":
        if xcnt==1:
            print(0)
        else:
            sta=(staxm-pow(2,n-i-1,mi))%mi
            print(1+calc(sta,0))
    else:
        sta=(staxp+pow(2,n-i-1,pl))%pl
        print(1+calc(sta,0))