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 Java (OpenJDK 11.0.6) (judge verdict: Runtime Error)

import java.util.*;
import java.io.*;
public class Main
{
    public static long pow(long x,int e,long m)
    {
        long ans=1;
        while(e>0)
        {
            if(e%2!=0)
            {
                ans*=x;
                ans%=m;
                e--;
            }
            else
            {
                x=x*x;
                e/=2;
                x%=m;
            }
        }
        return ans;
    }
    public static int countSetBits(long ans)
    {
        int cntr=0;
        while(ans>0)
        {
            if((ans&1)==1)
            cntr++;
            ans>>=1;
        }
        return cntr;
    }
    public static int steps(long ans)
    {
        int bits=countSetBits(ans);
        if(bits==0)
        return 0;
        return 1+steps(ans%bits);
    }
    public static void main(String args[])throws Exception
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw=new PrintWriter(System.out);
        int n=Integer.parseInt(br.readLine());
        char carr[]=br.readLine().toCharArray();
        int o=0;
        for(int i=0;i<n;i++)
        if(carr[i]=='1')o++;
        int m1=o-1,m2=o+1;
        long ans1=0,ans2=0;
        for(int i=n-1;i>=0;i--)
        {
            if(m1>0)
            {
                ans1+=(carr[i]-'0')*pow(2,n-1-i,m1);
                ans1%=m1;
            }
            ans2+=(carr[i]-'0')*pow(2,n-1-i,m2);
            ans2%=m2;
        }
        for(int i=0;i<n;i++)
        {
            long ans=0;
            if(carr[i]=='0')
            {
                ans=(ans2+pow(2,n-1-i,m2))%m2;
                ans=1+steps(ans);
            }
            else
            {
                ans=(ans1-pow(2,n-1-i,m1)+m1)%m1;
                ans=1+steps(ans);
            }
            pw.println(ans);
        }
        pw.flush();
        pw.close();
    }
}