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 {
	
	private static Scanner sc;
	
	public static void main (String args[]) {
		sc = new Scanner(System.in);
		int n = Integer.parseInt(sc.next());
		String x = sc.next();
		//n回回すのは確定
		for(int i = 1; i<=n; i++) {
			//ビット反転
			String changedSt = createChangedSt(x, i);
			//changedStの1の数
			long countOne = countOne(changedSt);
			//10進数に変換
			long xi  = Integer.parseInt(changedSt, 2); 
			//f(xi)を求める
			boolean b = true;
			int result = 0;
			while (b) {
				//xiをpopcount(xi)で割る popcount(xi) = countOne
				long der = xi%countOne;
				String strTen =  Long.toBinaryString(der);
				long countOneN = countOne(strTen);
				if (countOne==0) {
					b = false;
					System.out.println(result);
				} else {
					result++;
					xi=der;
					countOne=countOneN;
				}
			}
		}
	}

	private static String createChangedSt(String x, int i) {
		String stFirst = x.substring(0, i);
		String st = x.substring(i, i+1);
		String stLast = x.substring(i+1, x.length()-1);
		String changedSt;
		if("0".equals(st)) {
			changedSt = stFirst + "1" + stLast;
		} else {
			changedSt = stFirst + "0" + stLast;
		}
		return changedSt;
	}
		
		private static long countOne(String str) {
			return str.chars().filter(ch -> ch == '1').count();
		}
		
		

}