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)

N = int(input())
x = list(input())[::-1]

num = 0
pop_num = 0
for i in range(N):
    if x[i] == "1":
        num += 2 ** i
        pop_num += 1

def popcnt(n):
    c = (n & 0x5555555555555555) + ((n>>1) & 0x5555555555555555)
    c = (c & 0x3333333333333333) + ((c>>2) & 0x3333333333333333)
    c = (c & 0x0f0f0f0f0f0f0f0f) + ((c>>4) & 0x0f0f0f0f0f0f0f0f)
    c = (c & 0x00ff00ff00ff00ff) + ((c>>8) & 0x00ff00ff00ff00ff)
    c = (c & 0x0000ffff0000ffff) + ((c>>16) & 0x0000ffff0000ffff)
    c = (c & 0x00000000ffffffff) + ((c>>32) & 0x00000000ffffffff)
    return c
x = x[::-1]
import sys
sys.setrecursionlimit(10**10)

def calc(i,depth,n):
    if depth == 0:
        if x[i] == "1":
            tmp = (num - 2 ** (N - i)) % (pop_num - 1)
            return calc(i,depth+1,tmp)
        else:
            tmp = (num + 2 ** (N - i)) % (pop_num + 1)
            return calc(i,depth+1,tmp)
    else:
        if n == 0: return depth
        return calc(i,depth+1, n % popcnt(n))
        

for i in range(N):
    print(calc(i,0,0))