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)

import math
#import numpy as np
import itertools
import queue
import bisect
from collections import deque,defaultdict
import heapq as hpq
from sys import stdin,setrecursionlimit
#from scipy.sparse.csgraph import dijkstra
#from scipy.sparse import csr_matrix
ipt = stdin.readline
setrecursionlimit(10**7)

def popcorn(i,k):
    num = 0
    ii = i
    while ii > 0:
        if ii&1:
            num += 1
        ii //= 2
    if i%num:
        return popcorn(i%num,k+1)
    else:
        return k+1

def main():
    n = int(ipt())
    x = list(map(int,input()))
    si = sum(x)
    su = si+1
    sd = si-1
    amu = 0
    amd = 0
    n2u = [1]
    if si == 2:
        n2d = [0]
    else:
        n2d = [1]
    for i in x[::-1]:
        if i:
            amu = (amu+n2u[-1])%su
            amd = (amd+n2d[-1])%sd
        n2u.append((n2u[-1]*2)%su)
        n2d.append((n2d[-1]*2)%sd)
    for i,xi in enumerate(x):
        if xi:
            ni = (amd-n2d[-2-i])%sd
            if si == 1:
                print(0)
            elif ni == 0:
                print(1)
            else:
                print(popcorn(ni,1))
        else:
            ni = (amu+n2u[-2-i])%su
            if ni == 0:
                print(1)
            else:
                print(popcorn(ni,1))


    return None

if __name__ == '__main__':
    main()