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 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.*;
import java.io.*;
import java.math.BigInteger;
import static java.lang.Math.*;

public class Main {
    static int[] list;
    public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		
		// 入力
		int n = sc.nextInt();
		String x = sc.next();
		
		// 計算
		//int result = 0;
		int cnt = 0;
		for(int i = 0; i < n; i++){
		    if(x.charAt(i) == '1') cnt++;
		}
		list = new int[n+1];
		Arrays.fill(list, -1);
		list[0] = 0;
		list[1] = 1;
		for(int i = n; i >= 0; i--){
		    f(i);
		}
		/*
		for(int i = 0; i < 24; i++){
		    System.out.println(list[i]);
		}*/
		for(int i = 0; i < n; i++){
		    int result = 0;
		    char[] c = x.toCharArray();
		    int count = cnt;
		    if(c[i] == '0'){
		        c[i] = '1';
		        count++;
		    }else{
		        c[i] = '0';
		        count--;
		    }
		    String s = new String(c);
		    BigInteger bi = new BigInteger(s, 2);
		    if(!BigInteger.ZERO.equals(bi)){
		        BigInteger bj = bi.mod(new BigInteger(Integer.toString(count)));
		        result = list[bj.intValue()] + 1;
		    }
		    System.out.println(result);
		}
		

		// 出力
		//System.out.println(result);
	}
	
	public static int f(int n) {
	    if(list[n] != -1) return list[n];
	    int m = Integer.bitCount(n);
	    list[n] = f(n%m)+1;
	    return list[n];
	}
	
	public static class Scanner {
		private BufferedReader br; private StringTokenizer tok;
		public Scanner(InputStream is) throws IOException { br = new BufferedReader(new InputStreamReader(is));}
		private void getLine() throws IOException { while(!hasNext()) tok = new StringTokenizer(br.readLine());}
		private boolean hasNext() { return tok != null && tok.hasMoreTokens();}
		public String next() throws IOException { getLine(); return tok.nextToken();}
		public int nextInt() throws IOException { return Integer.parseInt(next());}
		public long nextLong() throws IOException { return Long.parseLong(next());}
		public double nextDouble() throws IOException { return Double.parseDouble(next());}
	}
	
}