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: Runtime Error)

def popcount(n):
    return str(bin(n)).count('1')
def f(n):
    if n==0:
        return 0
    else:
        return f(n%popcount(n))+1
N=int(input())
X=input()
K=X.count("1")
#k桁目がone,zeroのとき
ONE=0#K-1
ZERO=0#K+1
tmp1=1
tmp0=1
for i in range(N):
    if X[-i-1]=="1":
        ONE+=tmp1
        ONE%=(K-1)
        ZERO+=tmp0
        ZERO%=(K+1)
    tmp1*=2
    tmp1%=(K-1)
    tmp0*=2
    tmp0%=(K+1)
for i in range(N):
    nextnumber=-1
    if X[i]=="1":
        nextnumber=(ONE-pow(2,N-i-1,K-1))%(K-1)
    else:
        nextnumber=(ZERO+pow(2,N-i-1,K+1))%(K+1)
    print(f(nextnumber)+1)