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: Wrong Answer)

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        @SuppressWarnings("resource")
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        char[] x = sc.next().toCharArray();
        int c = 0;
        for (int i = 0 ; i < n ; i++) {
            if (x[i] == '1') {
                c++;
            }
        }
        int a = c - 1;
        int b = c + 1;
        int[] aa = new int[n];
        int[] ba = new int[n];
        int z = 1;
        if (a >= 1) {
            for (int i = 0 ; i < n ; i++) {
                z %= a;
                aa[i] = z;
                z *= 2;
            }
        }

        z = 1;
        for (int i = 0 ; i < n ; i++) {
            z %= b;
            ba[i] = z;
            z *= 2;
        }
        int d = 0;
        int e = 0;
        for (int i = 0 ; i < n ; i++) {
            if (x[i] == '1') {
                d += aa[n - i - 1];
                e += ba[n - i - 1];
            }
        }

        for (int i = 0 ; i < n ; i++) {
            int ans = 0;
            if (x[i] == '1') {
                int temp = d - aa[n - i - 1];
                int moda = a;
                while (temp > 0 && moda > 0) {
                    temp %= moda;
                    int temp2 = temp;
                    int cnt = 0;
                    while (temp2 > 0) {
                        if (temp2 % 2 == 1) {
                            cnt++;
                        }
                        temp2 /= 2;
                    }
                    moda = cnt;
                    ans++;
                }
            } else {
                int temp = e + ba[n - i - 1];
                int modb = b;
                while (temp > 0 && modb > 0) {
                    temp %= modb;
                    int temp2 = temp;
                    int cnt = 0;
                    while (temp2 > 0) {
                        if (temp2 % 2 == 1) {
                            cnt++;
                        }
                        temp2 /= 2;
                    }
                    modb = cnt;
                    ans++;
                }
            }
            System.out.println(ans);
        }


    }

}