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.Scanner;

public class Main {

	public static void main(String[] args) {
		long start = System.currentTimeMillis();
	    // 引数で標準入力System.inを指定する
	    Scanner scanner = new Scanner(System.in);
		try {

		    // 1行目を取得(a)
			int n  = scanner.nextInt();
			scanner.nextLine();
			String x  = scanner.next();
			scanner.nextLine();

			for (int i = 0; i < n; i++) {
				long  val = createVal(x, i);
				long  div = countLetter (Long.toBinaryString(val), '1');
				int count = 0;
				while (true) {
					count++;
					val = popcount(val, div);
					if(val == 0) {
						break;
					}
					div = countLetter (Long.toBinaryString(val), '1');
				}
				System.out.println(count);
			}


		} finally {
		    // Scannerクラスのインスタンスをクローズ
		    scanner.close();
		}
		long end = System.currentTimeMillis();
		// System.out.println((end - start) );
	}


	private static long createVal(String x, int keta) {

//		String baseBy = Integer.toBinaryString(base);

		StringBuilder sb = new StringBuilder();
		for (int i = 0; i < x.length(); i++) {
			if (i == keta) {
				if (x.charAt(i) == '0') {
					sb.append('1');
					continue;
				}
				sb.append('0');
				continue;
			}
			sb.append(x.charAt(i));
		}

		return Long.parseLong(sb.toString(), 2);
	}

	private static int countLetter(String value,  char target) {

		int count = 0;

		for (int i = 0; i < value.length(); i++) {
			if (value.charAt(i) == target) {
				count++;
			}
		}

		return count;
	}


	private static long popcount(long val, long div) {
		return val % div;
	}

	// ************************************
	// Common
	// ************************************

	private static final int[] changeArrayStringToInt(String[] strArray) {

		int[] intArray = new int[strArray.length];

		for (int i = 0; i < intArray.length; i++) {
			intArray[i] = Integer.parseInt(strArray[i]);
		}
		return intArray;
	}

}