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 f(n):
    cnt = 0
    while n:
        buf = n
        pc = 0
        while buf:
            pc += buf%2
            buf //= 2
        n = n%pc
        cnt += 1
    return cnt

N = int(input())
X = input()

v1, v2 = 0, 0
cnt = X.count("1")
for i, j in enumerate(X[::-1]):
    if j=="1":
        v1 += pow(2, i, cnt+1)
        v1 %= (cnt+1)

        if cnt>1:
            v2 += pow(2, i, cnt-1)
            v2 %= (cnt-1)

for i in range(N):
    p = N-i-1
    buf = 0
    c2 = 0
    if X[i]=="0":
        c2 = cnt+1
        buf = (v1+pow(2, p, c2))%c2
        print(f(buf)+1)
    else:
        c2 = cnt-1
        if c2>0:
            buf = (v2-pow(2, p, c2))%c2
            print(f(buf)+1)
        else:
            print(0)