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)

def readinput():
    n=int(input())
    x=input()
    return n,x

poptbl=[0]*(2*10**5+1)
def popcount(x):
    global poptbl
    if poptbl[x]!=0:
        return poptbl[x]
    
    poptbl[x]=bin(x).count('1')
    return poptbl[x]

ftbl=[0]*(2*10**5+1)
def f(x):
    global ftbl

    if ftbl[x]!=0:
        return ftbl[x]
    
    #print('f(x), x: {}'.format(x))
    if x==0:
        return 0
    else:
        ans=f(x%popcount(x))+1
        #print(ans)
        ftbl[x]=ans
        return ans    

def main():
    n=int(input())
    x=input()
    m=x.count('1')
    #print('m: {}'.format(m))
    xint=int(x,2)
    #print('xint: {}'.format(xint))
    if m!=1:
        xmodm1=xint%(m-1)
    else:
        xmodm1=1
    xmodp1=xint%(m+1)
    ans=[]
    b=2**(n-1)
    for i in range(n):
        if x[i]=='1':
            if m==1:
                a=0
                ans.append(a)
            else:
                xx=( xmodm1 + (m-1) -b%(m-1) )%(m-1)
                #ans.append(f(xx)+1)
                a=f(xx)+1
                ans.append(a)
        else:
            xx=( xmodp1 + b%(m+1) )%(m+1)
            #ans.append(f(xx)+1)
            a=f(xx)+1
            ans.append(a)
        b=b//2
        print(a)
    return ans

if __name__=='__main__':
    #n,x=readinput()
    ans=main()
    #for a in ans:
    #    print(a)