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: Time Limit Exceeded)

import sys
from math import ceil, floor, sqrt, sin, cos, pi
from itertools import accumulate, permutations, combinations
from fractions import gcd # 最大公約数
from collections import deque, Counter
from operator import itemgetter
from heapq import heappop,heappush
sys.setrecursionlimit(10**7)
def lcm(x, y): return ((x * y) // gcd(x, y)) # 最小公倍数
# list(map(int, input().split()))

n = int(input())
x = input()
s = ''
ans = [0 for _ in range(2**n + 1)]
pop = 0
for i in x:
    if i == '1': pop += 1

for i in range(n-1, -1, -1):
    if x[i] == '0':
        s = x[:i] + '1' + x[i+1:]
        pc = pop + 1
    else:
        s = x[:i] + '0' + x[i+1:]
        pc = pop - 1
    num = int(s, 2)
    cnt = 0
    while True:
        if num == 0:
            ans[int(s, 2)] = cnt
            break
        if ans[num] > 0:
            ans[int(s, 2)] = cnt + ans[num]
            break
        num = (num%pc)
        cnt += 1
        t = str(bin(num))
        pc = 0
        for j in t:
            if j == '1': pc += 1
for i in range(n):
    if x[i] == '0':
        s = x[:i] + '1' + x[i+1:]
    else:
        s = x[:i] + '0' + x[i+1:]
    print(ans[int(s, 2)])