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: Accepted)

import java.io.*;
import java.util.*;

public class Main{
	static long fastpow(long n,long ti,int mod) {
		 if (ti == 0) 
		        return 1l;
		if(ti%2==0) {
			long y=fastpow(n, ti/2,mod);
			long k=y*y;
			k%=mod;
			return k;
		}
		else {
			long y=fastpow(n, ti/2,mod);
			long k=((n*y)%mod)*y;
			k%=mod;
			return k;
		}
	}
	static void main() throws Exception{
		int n=sc.nextInt();
		char[]cs=sc.nextLine().toCharArray();
		int[]bits=new int[n];
		int cnt=0;
		for(int i=0;i<n;i++) {
			bits[i]=cs[i]-'0';
			cnt+=bits[i];
		}
		if(cnt==0) {
			for(int i=0;i<n;i++) {
				pw.println(1);
			}
			return;
		}
		long nummod=0,nummodG=0,nummodS=0;
		for(int i=0,pow=n-1;i<n;i++,pow--) {
			if(bits[i]==0)continue;
//			nummod=(nummod+fastpow(2, pow, cnt))%cnt;
			nummodG=(nummodG+fastpow(2, pow, cnt+1))%(cnt+1);
			if(cnt!=1)
				nummodS=(nummodS+fastpow(2, pow, cnt-1))%(cnt-1);
		}
		for(int i=0,pow=n-1;i<n;i++,pow--) {
			long curNum=0;
			int ans=1;
			if(bits[i]==0) {
				curNum=(nummodG+fastpow(2, pow, cnt+1))%(cnt+1);
				
			}
			else {
				if(cnt==1) {
					curNum=0;
					ans=0;
				}
				else
					curNum=(cnt-1+nummodS-fastpow(2, pow, cnt-1))%(cnt-1);
			}
			while(curNum>0) {
				curNum=(curNum%Long.bitCount(curNum));
				ans++;
			}
			pw.println(ans);
		}
	}
	public static void main(String[] args) throws Exception{
		pw=new PrintWriter(System.out);
		sc = new MScanner(System.in);
		int tc=1;
//		tc=sc.nextInt();
		while(tc-->0)
			main();
		pw.flush();
	}
	static PrintWriter pw;
	static MScanner  sc;
	static class MScanner {
		StringTokenizer st;
		BufferedReader br;
		public MScanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}
 
		public MScanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}
 
		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public int[] intArr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public long[] longArr(int n) throws IOException {
	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public int[] intSortedArr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        shuffle(in);
	        Arrays.sort(in);
	        return in;
		}
		public long[] longSortedArr(int n) throws IOException {
	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        shuffle(in);
	        Arrays.sort(in);
	        return in;
		}
		public Integer[] IntegerArr(int n) throws IOException {
	        Integer[]in=new Integer[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public Long[] LongArr(int n) throws IOException {
	        Long[]in=new Long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public String nextLine() throws IOException {
			return br.readLine();
		}
 
		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
 
		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
 
		public char nextChar() throws IOException {
			return next().charAt(0);
		}
 
		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}
 
		public boolean ready() throws IOException {
			return br.ready();
		}
 
		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
	static void shuffle(int[]in) {
		for(int i=0;i<in.length;i++) {
			int idx=(int)(Math.random()*in.length);
			int tmp=in[i];
			in[i]=in[idx];
			in[idx]=tmp;
		}
	}
	static void shuffle(long[]in) {
		for(int i=0;i<in.length;i++) {
			int idx=(int)(Math.random()*in.length);
			long tmp=in[i];
			in[i]=in[idx];
			in[idx]=tmp;
		}
	}
}