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: Time Limit Exceeded)

import sys
import copy
from collections import deque
import collections
stdin = sys.stdin

mod = 10**9+7

ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
ns = lambda: stdin.readline().rstrip()  # ignore trailing spaces

n = ni()
x = ns()
dp = {}


def popcount(n):
    b = bin(n)
    return b.count('1')

def f(xx):
    res = 0
    x = int(xx, 2)
    p = int(xx, 2)
    tmpdp = {}
    while p>0:
        if p in dp:
            res += dp[p]
            break
        pop = popcount(p)
        p = p%pop
        res += 1
        tmpdp[p] = res
    dp[x] = res
    for i,j in tmpdp.items():
        dp[i] = res - j
    return res

for i in range(n):
    if x[i]=='0':
        tmp = x[:i] + '1' + x[i+1:]
        print(f(tmp))
    else:
        tmp = x[:i] + '0' + x[i+1:]
        print(f(tmp))