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.io.PrintWriter;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Main {
	//java11

//	static int[] fn;
	static Map<Integer, Integer> fn;
	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);

		int N = sc.nextInt();
		String X = sc.next();

//		fn = new int[(int)Math.pow(2, N)];
//		fn = new int[10000000];
		fn = new HashMap<>();
//		Arrays.fill(fn, -1);

		PrintWriter pw = new PrintWriter(System.out);
		for(int i=0; i<N; i++) {
			StringBuilder sb = new StringBuilder(X);
			//bit反転
			if(X.charAt(i) == '0') {
				sb.setCharAt(i, '1');
			}else {
				sb.setCharAt(i, '0');
			}

			int Xi = Integer.parseInt(sb.toString(), 2);

			pw.println(fn(Xi));

		}

		pw.flush();

	}

	public static int fn(int n) {
		if(fn.get(n) != null) {
			return fn.get(n);
		}

		if(n == 0) {
//			fn[n] = 0;
			fn.put(n, 0);
			return fn.get(n);
		}

		if(popcount(n) == 0) {
			fn.put(n, fn(0) + 1);
		}else {
			fn.put(n, fn(n % popcount(n)) + 1);
		}
		return fn.get(n);
	}

	public static int popcount(int n) {
		return Integer.bitCount(n);
	}

}