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 PyPy2 (7.3.0) (judge verdict: Wrong Answer)

import collections
n = int(raw_input())
x = int(raw_input(),2)
mem = {0:0}


def k(x, mod):
	ximod = 1 % mod
	for i in range(x+1):
		if i: ximod *= 2
		ximod %= mod
	return ximod

def g(x, mod):
	ximod = 1 % mod
	r = 0
	for i in range(n):
		
		if i: ximod *= 2
		ximod %= mod
		if x & (1 << i):
			r += ximod
			r %= mod
	return r
def f(x):
	if x in mem:
		return mem[x]
	cc=  collections.Counter(bin(x))
	compute = g(x,cc['1'])
	if compute == 0: 
		mem[x] = 1
		return 1
	mem[x] = 1 + f(compute)
	return 1 + f(compute)


for i in range(n - 1, -1, -1):
	if i == n - 1:
		t = f(x ^ (1 << i))
		c = collections.Counter(bin(x))['1'] - 1
		print t
	else:
		#k(i, c) == 0
		if  c and k(i, c) == 0:
			print t
		else:
			print f(x ^ (1 << i))