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 PyPy3 (7.3.0) (judge verdict: Time Limit Exceeded)

N = int(input())
x = input()

#N=3
#x="011"
def onecount(mojix):
    ones=0
    for j in mojix:
        if j=="1":
            ones+=1
    return ones


def popcount(X,ones,count):
    #print("popcount start",count,X,bin(X),ones)
    
    if ones==0:
        return count
    
    else:
        X=X%ones
        
        count+=1

        ones=onecount(str(bin(X)))
        
        return popcount(X,ones,count)

GroundOnes=onecount(x)
GroundX=int(x,2)

for i in range(N):
    
    if x[i]=="1":
        x_ith="0"
        ones=GroundOnes-1
        X=GroundX-2**(N-i-1)
    else:
        x_ith="1"
        ones=GroundOnes+1
        X=GroundX+2**(N-i-1)
            
    if i==0:
        new2bit=x_ith+x[i+1:]
    elif i<N-1:
        new2bit=x[:i]+x_ith+x[i+1:]
    else:
        new2bit=x[:i]+x_ith

    # print("for loop",i)
    # print("ones",ones)
    # print("new2bit",new2bit)
    # print(" ------ return --- ",popcount(X,ones,0))
    print(popcount(X,ones,0))