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: Accepted)

N = int(input())
sX = input()

origin_count = sX.count('1')
one_r_count = origin_count -1
zero_r_count = origin_count +1

one_mod = 0
zero_mod =0
for b in sX:
    if one_r_count != 0:
        one_mod = (one_mod * 2 + int(b) )% one_r_count
    zero_mod = (zero_mod * 2  + int(b) )% zero_r_count

f = [0] * 220000
pop = [0]* 220000

for i in range(1,220000):
    pop[i] = pop[i//2] + i % 2
    f[i] = f[i % pop[i]] + 1

onepow = [1] * 220000
zeropow = [1]* 220000

for i in range(1,220000):
    if one_r_count != 0:
        onepow[i] = onepow[i-1] * 2 % one_r_count
    zeropow[i] = zeropow[i-1] * 2 % zero_r_count

for i in range(1,N+1):
    if sX[i-1] == '1':
        if one_r_count != 0:
            Xi_n = one_mod - onepow[N-i]
            Xi_n %= one_r_count
            print(f[Xi_n]+1)
        else:
            print(0)
    if sX[i-1] == '0':
        Xi_n = zero_mod + zeropow[N-i]
        Xi_n %= zero_r_count
        print(f[Xi_n]+1)