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.*;



public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);
    int n = scanner.nextInt();
    scanner.nextLine();
    String s_x = scanner.nextLine();
    char[] c_x = s_x.toCharArray();

    for(int i=0; i<n; i++){
      char[] tmp = new char[n];
      System.arraycopy(c_x, 0, tmp, 0, n);
      tmp[i] = (char)((Integer.parseInt(String.valueOf(tmp[i])) + 1) % 2 + '0');
      String x = new String(tmp);
      int ans = f(x,0);
      System.out.println(ans);

    }
    
   
    
    scanner.close();
    return;
  }

  public static int f(String x, int ans){
    int dec = Integer.parseInt(x, 2);
    // System.out.println(x +"," + dec+","+ans);
    if(dec != 0){
      int pop = 0;
      for(char c_x : x.toCharArray() ){
        if(c_x == '1'){
          pop++;
        }       
      }
      dec = dec % pop;
      ans += 1;
      ans = f(Integer.toBinaryString(dec), ans);
    }
    return ans;
  }
  
}