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.util.*;

public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.next());
        String s = sc.next();
        char[] x = s.toCharArray();
        int popcount = 0;
        for(int i=0; i<n; i++){
            if(x[i]=='1'){
                popcount++;
            }
        }

        //0 -> popcount-1
        //2 -> popcount+1
        int[] remx = new int[3];
        int[][] rem2beki = new int[3][n+1];
        for(int i=0; i<3; i+=2){
            int mod = popcount-1+i;
            if(mod<=0){
                continue;
            }
            rem2beki[i][0] = 1;
            for(int j=0; j<n; j++){
                rem2beki[i][j+1] = rem2beki[i][j]*2%mod;
                remx[i] = remx[i]*2 + ((x[j]=='1')?1:0);
                remx[i] %= mod;
            }
        }

        for(int i=0; i<n; i++){
            if(popcount==1 && x[i]=='1'){
                System.out.println(0);
                continue;
            }

            int j = (x[i]=='0') ? 2 : 0;
            int res = remx[j];
            if(x[i]=='0'){
                //mod popcount+1
                res += rem2beki[j][n-1-i];
                res %= popcount+1;
            }else{
                //mod popcount-1
                res -= rem2beki[j][n-1-i] - (popcount-1);
                res %= popcount-1;
            }

            System.out.println(calc(res)+1);
        }
    }

    public static int calc(int n){
        if(n==0){
            return 0;
        }

        int popcount = Integer.bitCount(n);
        int res = n%popcount;

        return calc(res)+1;
    }
}