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())

ans_count = [-1 for _ in range(n)]

for i in range(1,n+1):
    b = bin(i)
    b=list(b)
    a = b.count("1")
    f = i%a
    if f == 0:
        ans_count[i-1]=1
    else:
        ans_count[i-1] = ans_count[f-1]+1

ans_count = [0]+ ans_count


inp = input()
x = list(inp)

org = int(inp, 2)
if "1" in x:      
    l = x.count("1")
else:
    l = 0





for i in range(n):
    a = x[i]
    print(0)
#     if a=="0":
#         N = org+(2**(n-i-1))
#         m= N%(l+1)
#         print(ans_count[m]+1)
#     else:
#         if l ==1:
#             print(0)
#         else:
#             N = org-(2**(n-i-1))
#             m=N%(l-1)
#             print(ans_count[m]+1)