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)

from functools import lru_cache

import sys

sys.setrecursionlimit(4100000)

n = int(input())
x = input()
num = int(x, 2)


def popcnt(n):
    c = (n & 0x5555555555555555) + ((n >> 1) & 0x5555555555555555)
    c = (c & 0x3333333333333333) + ((c >> 2) & 0x3333333333333333)
    c = (c & 0x0F0F0F0F0F0F0F0F) + ((c >> 4) & 0x0F0F0F0F0F0F0F0F)
    c = (c & 0x00FF00FF00FF00FF) + ((c >> 8) & 0x00FF00FF00FF00FF)
    c = (c & 0x0000FFFF0000FFFF) + ((c >> 16) & 0x0000FFFF0000FFFF)
    c = (c & 0x00000000FFFFFFFF) + ((c >> 32) & 0x00000000FFFFFFFF)
    return c


# aは1の数


@lru_cache(None)
def calc2(a):
    if a == 0:
        return 0
    return calc2(a % popcnt(a)) + 1


ans = [0] * n
cnt = x.count("1")

kotae = [0] * (cnt + 2)
for i in range(cnt + 1):
    kotae[i] = calc2(i)


for i in range(n):
    if x[i] == "1":
        temp = cnt - 1
    else:
        temp = cnt + 1
    t = 1 << (n - i - 1)

    if num ^ t:
        ans[i] = calc2(temp)
    else:
        ans[i] = 0
print(*ans, sep="\n")