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.util.*;
import java.lang.*;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n;
        String s;
        n = Integer.valueOf(in.nextLine());
        s = in.nextLine();

        int pop1, pop2;
        pop1 = popCount(s);
        pop2 = pop1-1;
        pop1++;

        long x1=0, x2=0;
        for(int i=0; i<n; i++){
            if(s.charAt(i)=='1'){
                x1 += Math.pow(2, n-i-1) % pop1;
                x2 += Math.pow(2, n-i-1) % pop2;
            }
        }

        for(int i=0;i<n;i++){
            String x = s.substring(0, i) + (s.charAt(i)=='1'?'0':'1') + s.substring(i+1);
            
            int count = 0;
            long xi = 0;
            int pop;
            
            if(s.charAt(i)=='1'){
                pop = pop2;
                xi = x2;
            }else{
                pop = pop1;
                xi = x1;
            }

            if(pop>0){
                if(s.charAt(i)=='1'){
                    xi -= Math.pow(2, n-i-1) % pop;
                }else{
                    xi += Math.pow(2, n-i-1) % pop;
                }
                xi = xi % pop;
                count++;
            }

            while(xi > 0){
                xi = xi % popCount(xi);
                count++;
            }
            print(count);
        }

        in.close();
    }

    private static int popCount(String s){
        int count = 0;
        for(int i=0; i<s.length(); i++){
            count += s.charAt(i)=='1' ? 1 : 0;
        }
        return count;
    }

    private static int popCount(long n){
        int count = 0;
        while(n>0){
            count += n%2;
            n /= 2;
        }
        return count;
    }

    private static void print(Object obj){
        System.out.println(obj);
    }
} // javac MyTest.java; java MyTest < input.txt > out