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 1.8.0) (judge verdict: Runtime Error)

import java.util.Scanner;

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

		for (int i = 1; i <= n; i++) {
			String flipX = flipBinStr(x, i);
			System.out.println(calcFn(Long.parseLong(flipX, 2)));
		}
	}

	static int popcount(long n) {
		char target = '1';
		int count = 0;
		String nBin = Long.toBinaryString(n);
		for (char x : nBin.toCharArray()) {
			if (x == target) {
				count++;
			}
		}
		return count;
	}

	static int calcFn(long n) {
		int count = 0;
		while (n != 0) {
			n = n % popcount(n);
			count++;
		}
		return count;
	}

	static String flipBinStr(String x, int i) {
		StringBuilder result = new StringBuilder(x);
		char target = result.charAt(i - 1);
		if (target == '1') {
			result.setCharAt(i - 1, '0');
		} else {
			result.setCharAt(i - 1, '1');
		}
		return result.toString();
	}
}