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.math.BigInteger;
import java.util.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		int[] map = new int[200001];
		int[] ppcount = new int[200001];
		map[0] = 0;
		map[1] = 1;
		ppcount[0] = 0;
		ppcount[1] = 1;
		for (int i = 2; i < 200001; i++) {
			ppcount[i] = ppcount[i/2] + i%2;
			map[i] = map[i%ppcount[i]] + 1;
		}

		Scanner sc = new Scanner(System.in);
		int l = sc.nextInt();
		String n = sc.next();

		char[] array = n.toCharArray();
		int count = 0;
		for (int i = 0; i < array.length;i++) {
			if (array[i] == '1') {
				count++;
			}
		}

		for (int i = 0; i < array.length;i++) {
			String string = null;
			int newCount = count;
			char[] newArray = Arrays.copyOf(array, array.length);
			if (array[i] == '1') {
				newArray[i] = '0';
				string = String.valueOf(newArray);
				newCount-=1;
			} else {
				newArray[i] = '1';
				string = String.valueOf(newArray);
				newCount+=1;
			}
			if (newCount == 0) {
				System.out.println(0);
			} else {
				System.out.println(map[binaryMod(string, newCount)] + 1);
			}
		}

	}

	private static int binaryMod(String binaryString, int mod) {
		if (mod == 0) {
			return -1;
		}
		int temp = 0;
		for (int i = 0; i < binaryString.length(); i++) {
			temp = temp % mod;
			temp = (temp * 2) + Integer.valueOf(String.valueOf(binaryString.charAt(i)));
		}
		return temp % mod;
	}

}