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: Wrong Answer)

import java.io.*;
import java.util.*;
public class Main
{
	
	private void solve()throws Exception
	{
		int n=nextInt();
		int bit[]=new int[n+1];
		String s=nextLine();
		int cnt=0;
		for(int i=1;i<=n;i++)
		{
			bit[i]=s.charAt(i-1)-'0';
			if(bit[i]==1)
				cnt++;
		}
		int d1=0;
		if(cnt!=1)
		{
			for(int j=n;j>=1;j--)
			{
				if(bit[j]==1)
					d1=(d1+(int)modpow(2,n-j,cnt-1))%(cnt-1);
			}
		}
		int d2=0;
		for(int j=n;j>=1;j--)
		{
			if(bit[j]==1)
				d2=(d2+(int)modpow(2,n-j,cnt+1))%cnt+1;
		}
		for(int i=1;i<=n;i++)
		{
			if(bit[i]==1 && cnt==1)
			{
				out.println(0);
				continue;
			}
			int d;
			if(bit[i]==1)
				d=(d1-(int)modpow(2,n-i,cnt-1)+(cnt-1))%(cnt-1);
			else
				d=(d2-(int)modpow(2,n-i,cnt+1)+(cnt+1))%(cnt+1);
			int ans=1;
			while(d>0)
			{
				d=d%Integer.bitCount(d);
				ans++;
			}
			out.println(ans);
		}
	}
	long modpow(long a,long b,long mod)
	{
	    long ret=1;
	    while(b!=0)
	    {
	        if(b%2==1)
	            ret=(ret*a)%mod;
	        a=(a*a)%mod;
	        b=b/2;
	    }
	    return ret;
	} 
	 
	///////////////////////////////////////////////////////////

	public void run()throws Exception
	{
		br=new BufferedReader(new InputStreamReader(System.in));
		st=null;
		out=new PrintWriter(System.out);

		solve();
		
		br.close();
		out.close();
	}
	public static void main(String args[])throws Exception{
		new Main().run();
	}
	BufferedReader br;
	StringTokenizer st;
	PrintWriter out;
	String nextToken()throws Exception{
		while(st==null || !st.hasMoreTokens())
		st=new StringTokenizer(br.readLine());
		return st.nextToken();
	}
	String nextLine()throws Exception{
		return br.readLine();
	}
	int nextInt()throws Exception{
		return Integer.parseInt(nextToken());
	}
	long nextLong()throws Exception{
		return Long.parseLong(nextToken());
	}
	double nextDouble()throws Exception{
		return Double.parseDouble(nextToken());
	}
}