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: Wrong Answer)

n=int(input())
x=input()
chk0=chk1=chk2=0
b=x.count("1")
if b==1 and x=="1":
    print(0)
    exit()
elif b==1 and x[-1]=="1":
    for i in range(n-1):
        print(2)
    print(0)
    exit()

for i,j in enumerate(x[::-1]):
    if j=="1":
        chk1+=pow(2,i,b+1)
        if b>1: chk2+=pow(2,i,b-1)
        else: chk2=0
l=[]
for i,j in enumerate(x[::-1]):
    if b: a1=pow(2,i,b+1)
    else: a1=1
    if b>1: a2=pow(2,i,b-1)
    else: a2=1
    if j=="0":
        t=chk1+a1
    else:
        if b>1: t=chk2+b-1-a2
        else: t=0
    ans=0
    f=1
    while t:
        if f:
            if j=="0": t%=(b+1)
            else: t%=(b-1)
            f=0
        else:
            t%=bin(t).count("1")
        ans+=1
    l.append(ans)
for i in l[::-1]:
    print(i)