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)

#!python3

import sys
iim = lambda: map(int, sys.stdin.readline().rstrip().split())

def resolve():
    N, S = sys.stdin.read().split()
    N = int(N)
    cnt0 = S.count("1")

    if cnt0 == 0:
        print(*[1]*N, sep="\n")
        raise 1
        return

    X = int(S, 2)

    def f2():
        cache = [-1] * max(9, cnt0+1)
        cache[0] = 0
        for i in range(4): cache[1<<i] = 1
        def f2(x):
            if cache[x] != -1:
                return cache[x]

            y = bin(x).count("1")
            z = 1 + f2(x % y)
            cache[x] = z

            return z
        f2.cache = cache
        return f2
    f2 = f2()

    dc = {}
    for k, v in (("0", 1), ("1", -1)):
        x = cnt0 + v
        y = X % x if x else 0
        dc[k] = (x, v, y, [1, N-1])

    ans = [0] * N
    for i in reversed(range(N)):
        si = S[i]
        cnt, v, n, m1 = dc[si]

        if cnt == 0:
            ans[i] = 0
            continue

        m2 = (m1[0]<<(m1[1]-i)) % cnt
        n = (n + v*m2) % cnt
        m1[0], m1[1] =  m2, i

        ans[i] = 1 + f2(n)

    print(*ans, sep="\n")


if __name__ == "__main__":
    resolve()