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: Time Limit Exceeded)

import java.util.*;
import java.io.*;
import java.math.BigInteger;

public class Main {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);

		int n = in.nextInt();
		String s = in.next();

		BigInteger x = BigInteger.ZERO;
		for (int i=0; i<n; i++)
			if (s.charAt(i) == '1')
				x = (x.multiply(BigInteger.TWO)).add(BigInteger.ONE);
			else
				x = x.multiply(BigInteger.TWO);

		for (int i=0; i<n; i++) {
			BigInteger y = x.xor(BigInteger.TWO.pow(n-1-i));

			int count = 0;
			while (y.compareTo(BigInteger.ZERO) != 0) {
				//System.out.println(y);
				y = y.remainder(new BigInteger("" + y.bitCount()));
				count++;
			}
			System.out.println(count);
		}
	}
}