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) {

		Scanner sc = new Scanner(System.in);
		String input1 = sc.nextLine();
		String input2 = sc.nextLine();
		StringBuilder str = new StringBuilder(input2);


		for (int i = 0; i < str.length(); i++) {
			// ビット反転
			str = changeNumber(new StringBuilder(input2), i);

			// 1の数カウント
			int count = countOne(str);

			// ２進数->１０進数
			String str2 = new String(str);
			int numTen = Integer.parseInt(str2, 2);

			int count2 = 0;

			while(numTen != 0) {
				numTen = numTen % count; // decは１０進数int
				count2++;

				//10 -> 2
				String numTwo = Integer.toBinaryString(numTen);

				//String -> Bulider
				count = countOne(new StringBuilder(numTwo));
				numTen = Integer.parseInt(numTwo, 2);

			}

			System.out.println(count2);


		}

	}

	static StringBuilder changeNumber(StringBuilder str, int start) {
		if(str.substring(start, start+1).equals("0")) {
			str.setCharAt(start, '1');
		}else {
			str.setCharAt(start, '0');
		}

		return str;

	}

	static int countOne(StringBuilder str) {
		int count = 0;
		for(int j = 0; j < str.length(); j++) {

			if(str.substring(j, j+1).equals("1")){
				count++;
			}
		}
		return count;
	}



}