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 void main(String[] args) throws IOException{
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));
		int n = Integer.parseInt(reader.readLine());
		char[] list = reader.readLine().toCharArray();
		int clip = 0;
		for(int i = 0;i<n;i++){
			if(list[i]=='1') clip++;
		}
		int[][] rems = new int[2][n+1];
		ArrayList<Integer> l = new ArrayList<>();
		l.add(-1);
		l.add(1);	
		for(int i=0;i<2;i++){
			int temp = clip + l.get(i);
			for(int j = 0;j<=n;j++){
				if(temp==0){
					rems[i][j] = 0;
					continue;
				}
				if(j==0) rems[i][j] = 1%temp;
				else rems[i][j] = (rems[i][j-1]*2)%temp;
			}
		}
		int[] remlist = new int[2];
		for(int i = 0;i<2;i++){
			int temp = clip + l.get(i);
			for(int j = 0;j<n;j++){
				if(list[j]=='1'){
					remlist[i] = (remlist[i]+rems[i][n-j-1])%temp; 
				}
			}
		}
		for(int i = 0;i<n;i++){
			if(list[i]=='1'){
				if(clip==1){
					writer.write(1+"\n");
					continue;
				}
				int lp = ((remlist[0] - rems[0][n-i-1]) + (clip-1))%(clip-1);
				writer.write((1+solve(lp))+"\n"); 
			}
			else{
				int lp = (remlist[1] + rems[1][n-i-1])%(clip+1);
				writer.write((1+solve(lp))+"\n");	
			}
		}
		writer.flush();
	}
	public static int solve(int p){
		if(p==0) return 0;
		int temp = p;
		int cnt = 0;
		while(p>0){
			if(p%2==1) cnt++;
			p=p>>1;
		}
		if(cnt==0) return 1; 
		return 1 + solve(temp%cnt);
	}
}