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)

def cin():
    return input().split()
def qpow(a,b,p):
    if p == 1 or p == 0:
        return 0
    a = (a%p+p)%p
    ans = 1
    while b!=0:
        if (b&1) != 0:
            ans = (ans * a) % p
        b >>= 1
        a = a * a  % p
    return ans

def f(n):
    if n == 0:
        return 0
    return f(n % bin(n).count("1")) + 1

n = int(input())
x = list(input())
x = x[::-1]
cnt = 0;l = len(x)
r1 = 0;r2 = 0
for i in x:
    if i == '1':
        cnt += 1
i = 0
while i < l:
    if x[i] == '1':
        r1 += qpow(2,i,cnt+1)
        r2 += qpow(2,i,cnt-1)
    i += 1
i = 0
a = []
while i < l:
    if x[i] == '0':
        t = f((r1+qpow(2,i,cnt+1))%(cnt+1))+ 1
    else:
        if cnt-1 !=  0:
            t = f((r2-qpow(2,i,cnt-1)+(cnt-1))%(cnt-1)) + 1
        else:
            t = 0
    i += 1
    a.append(t)
a = a[::-1]
for i in a:
    print(i)