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: Time Limit Exceeded)

import java.util.Scanner;

class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = Integer.parseInt(sc.next());
    String xdata = sc.next();
    sc.close();

    for (int i=0; i<n; i++) {
    	StringBuffer x = new StringBuffer(xdata);

    	if (x.substring(i,i+1).equals("1"))
    		x.setCharAt(i, '0');
    	else x.setCharAt(i, '1');

    	int cnt = 0;
    	int f = binToDec(x);
    	while (f != 0) {
    		int p = popCount(x);
    		f = func(f, p);
    		x = decToBin(n, f);
    		cnt ++;
    	}
    	System.out.println(cnt);
    }


  }
  public static int popCount(StringBuffer n) {
	  int pop=0;
	  for (int i=0; i<n.length(); i++) {
		  if (n.substring(i,i+1).equals("1")) pop++;
	  }
	  return pop;
  }

  public static int func(int n, int p) {
	  int f=0;
	  f = n%p;
	  return f;
  }

  public static StringBuffer decToBin (int n, int f) {
	  StringBuffer x = new StringBuffer();
	  for (int i=n; i>=0; i--) {
		  x.append((f%(int)Math.pow(2, i+1))/(int)Math.pow(2, i));
	  }
	  return x;
  }

  public static int binToDec (StringBuffer x) {
	  int f=0;
	  for (int i=0; i<x.length(); i++) {
		  if (x.substring(i,i+1).equals("1"))
			  f += (int)Math.pow(2, x.length()-i-1);
	  }
	  return f;
  }
}