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)

def popcnt1(n):
    return bin(n).count("1")

def okikae(n):
    kaisu = 0
    if n == 0:
        return 0
    while 1:
        P = popcnt1(n)
        n %= P
        kaisu += 1
        if n == 0:
            return kaisu

OKI = [okikae(i) for i in range(200001)]
X = int(input())
Norig = input()
N = int(Norig,2)
Npop = Norig.count("1")
F = Npop + 1
H = Npop - 1
if H == 0:
    H = -1
FueAmari = N%(Npop + 1)
if Npop != 1:
    HeruAmari = N%(Npop - 1)
else:
    HeruAmari = -1
Fues = []
Herus = []
fue = 1
heru = 1
for i in range(2*(10**5+1)):#2^0の桁から
    fue %= F
    heru %= H
    Fues.append(fue)
    Herus.append(heru)
    fue *= 2
    heru *= 2
RNorig = reversed(Norig)
Results = []
Zougen = []#増えたらTrue
Kaisu = [0] * X
#上の桁から見ていく。ゼロなら減ってイチなら増える
for n,bit in enumerate(Norig):#2**(N-1-n)の桁
    if bit == "0":#増える
        Zougen.append(True)
        print(1+OKI[(FueAmari + Fues[X-1-n] )% F])
    else:
        if H == -1:
            print(0)
            continue
        Zougen.append(False)
        print(1+OKI[(HeruAmari - Herus[X-1-n]) % F])