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 C (GCC 9.2.1) (judge verdict: Runtime Error)

#include <stdio.h>
#include <string.h>
void rev_string(int str[],int len)
{
    int i;
    for (i = 0; i < len / 2; i++) {
        int temp = str[i];
        str[i] = str[len-i-1];
        str[len-i-1] = temp;
    }
}

int main(void){
    // Your code here!
    int n;
    scanf("%d",&n);
    char b[n+10];
    int a[n+1];
    scanf("%s",b);
    int ac[n+1];
    for(int i=0;i<n+1;i++)
    {
        if(b[i]=='0')
        a[i]=0;
        else a[i]=1;
    }
    for(int i=0;i<n;i++)    
        {if(i!=0)
            {
            a[i-1]=(a[i-1]+1)%2;
            }
            a[i]=(a[i]+1)%2;
         
            for(int i=0;i<n+1;i++)
            {
                ac[i]=a[i];
            }
            int q=n;
            int ij=0;
            while(1==1)
            {  
            ij++;
            int k=0;
            int l=0;
            for(int j=0;j<q;j++)
                {
                k=k*2;
                k=k+ac[j];
                if(ac[j]==1)
                l++;
                }
            if(k%l==0)
                {
                printf("%d\n",ij);
                break;
                }
            int c=k%l;
                for(int i=0;i<n;i++)
                {
                    ac[i]=c%2;
                    c=c/2;
                    if(c==0)
                    {
                        q=i+1;
                        break;
                    }
                }
            rev_string(ac,q);
	
            }
        }
    return 0;
}