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)

n = int(input())
X = input()
x = int(X,2)
bitx = bin(x).count('1')
 
ans = [0]*(bitx+2)
for i in range(1,bitx+2):
    ans[i] = 1 + ans[i%bin(i).count('1')]
#print(ans)
l = []
bitp = bitx + 1
bitm = bitx - 1
xmp = 0
xmm = 0

for i in range(n):
    if X[i] == '1':
        xmp += pow(2, n-(i+1), bitp)
        xmm += pow(2, n-(i+1), bitm)
if bitx != 1:
    for i in range(n):
        if X[i] == '1':
            xm = (xmm - pow(2,n-i-1,bitm)) % bitm
            l.append(str(1+ans[xm]))
        else:
            xm = (xmp + pow(2,n-i-1,bitp))%bitp
            l.append(str(1+ans[xm]))

else:
    for i in range(n):
        if X[i] == '1':
            l.append(str(0))
        else:
            xm = (xmp + pow(2,n-i-1,bitp))%bitp
            l.append(str(1+ans[xm]))

print('\n'.join(l))