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: Wrong Answer)

n = int(input())
x = input()
pc = 0
for i in range(len(x)):
    if x[i]=='1':
        pc += 1
def popcount(s):
    c = 0
    for i in range(len(s)):
        if s[i]=='1':
            c += 1
    return c
d = dict()
for i in range(len(x)):
    X = x[0:i] + ('0' if x[i]=='1' else '1') + x[i+1:len(x)]
    if x[i]=='1':
        pop = pc-1
    else:
        pop = pc+1
    v = int('0b'+X, 0)
    if pop==0:
        print('0')
        continue
    new = v % pop
    if new == 0:
        print('1')
        continue
    counter = 1
    while True:
        counter += 1
        oldnew = new
        new = new % popcount(str(bin(new)))
        if new in d:
            d[oldnew] = d[new] + 1
            print(counter + d[new])
            break
        if new == 0:
            d[oldnew] = 0
            print(counter)
            break