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: Accepted)

def popcount(u):
    t = bin(u)
    h = t.count("1")
    return u % h 

n = int(input())
x = list(input())
c = x.count("1") + 1
d = c - 2

amaric = [1 % c]
if d >= 1:
    amarid = [1 % d]
for i in range(n-1):
    amaric.append((amaric[-1]*2)%c)
    if d >= 1:
        amarid.append((amarid[-1]*2)%d)

motoc = 0
motod = 0
for i in range(n):
    if x[i] == "1":
        motoc += amaric[n-1-i]
        if d >= 1:
            motod += amarid[n-1-i]

for i in range(n):
    if x[i] == "0":
        op = motoc
        op += amaric[n-1-i]
        op %= c
    elif x[i] == "1":
        if d >= 1:
            op = motod
            op -= amarid[n-1-i]
            op %= d
        elif d == 0:
            print(0)
            continue
    cou = 1
    while True:
        if op == 0:
            break
        op = popcount(op)
        cou += 1
    print(cou)