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: Runtime Error)

n = int(input())
x = list(input())

dp=[-1]*(n+1)

dp[0]=0
dp[1]=1
co=[-1]*(n+1)

def fun(num,step):
    num2=num%sum(list(map(int,bin(num)[2:])))
    if dp[num2]!=-1:
        dp[num]=dp[num2]+step
    else:
        fun(num2,step+1)

for i in range(2,n//2):
    fun(i,1)

num1=''.join(x)
num2=int(num1,2)
su = sum(list(map(int,x)))

for i in range(n):
    num4=num2
    su2=su
    if x[i]=='1':
        num4-=pow(2,(n-1-i))
        su2-=1
    else:
        num4+=pow(2,(n-1-i))
        su2+=1
    if su2 == 0:
        print('0')
    elif num4<n//2:
        print(dp[num4])
    else:
        num3=num4%su2
        if num3<n//2:
            print(dp[num3]+1)
        else:
            num5=num3%sum(list(map(int,bin(num3)[2:])))
            print(dp[num5]+2)