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: Runtime Error)

# -*- coding: utf-8 -*-
import sys
import math
import os
import itertools
import string
import heapq
import _collections
from collections import Counter
from collections import defaultdict
from collections import deque
from functools import lru_cache
import bisect
import re
import queue
import copy
import decimal


class Scanner():
    @staticmethod
    def int():
        return int(sys.stdin.readline().rstrip())

    @staticmethod
    def string():
        return sys.stdin.readline().rstrip()

    @staticmethod
    def map_int():
        return [int(x) for x in Scanner.string().split()]

    @staticmethod
    def string_list(n):
        return [Scanner.string() for i in range(n)]

    @staticmethod
    def int_list_list(n):
        return [Scanner.map_int() for i in range(n)]

    @staticmethod
    def int_cols_list(n):
        return [Scanner.int() for i in range(n)]


def pop_count(x):
    x = x - ((x >> 1) & 0x5555555555555555)
    x = (x & 0x3333333333333333) + ((x >> 2) & 0x3333333333333333)
    x = (x + (x >> 4)) & 0x0f0f0f0f0f0f0f0f
    x = x + (x >> 8)
    x = x + (x >> 16)
    x = x + (x >> 32)
    return x & 0x0000007f


def f(x):
    if x == 0:
        return 0
    return f(x % pop_count(x)) + 1


def solve():
    N = Scanner.int()
    S = Scanner.string()
    pc = sum([s == '1' for s in S])
    ans = [0 for _ in range(N)]
    for b in range(2):
        npc = pc
        if b == 0:
            npc += 1
        else:
            npc -= 1
        r0 = 0
        for i in range(N):
            r0 = ((r0 * 2) + int(S[i])) % npc
        k = 1
        for i in reversed(range(N)):
            if int(S[i]) == b:
                r = r0
                if b == 0:
                    r = (r + k) % npc
                else:
                    r = (r - k + npc) % npc
                ans[i] = f(r) + 1
            k = (k * 2) % npc
    print(*ans, sep='\n')


def main():
    sys.setrecursionlimit(1000000)
    # sys.stdin = open("sample.txt")
    # T = Scanner.int()
    # for _ in range(T):
    #     solve()
    # print('YNeos'[not solve()::2])
    solve()


if __name__ == "__main__":
    main()