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.io.PrintWriter;
import java.util.*;

public class Main {
    void run() {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

        int n = sc.nextInt();
        String s = sc.next();
        int m = 0;
        for (int i = 0; i < n; i++) {
            if (s.charAt(i) == '1') m++;
        }

        int[] maArr = new int[n + 1];
        int[] mbArr = new int[n + 1];
        maArr[0] = 1;
        mbArr[0] = 1;
        // 2のi乗の累積
        for (int i = 0; i < n; i++) {
            maArr[i + 1] = (maArr[i] * 2) % (m + 1);
            mbArr[i + 1] = (mbArr[i] * 2) % (m - 1);
        }

        int maMod = 0;
        int mbMod = 0;
        for (int i = 0; i < n; i++) {
            if (s.charAt(i) == '1') {
                maMod = (maMod + maArr[n - i - 1]) % (m + 1);
                mbMod = (mbMod + mbArr[n - i - 1]) % (m - 1);
            }
        }

        int[] modArr = new int[m + 1];

        for (int bit = 1; bit < m + 1; bit++) {
            int count = 0;
            for (int i = 0; i < 18; i++) {
                if ((bit & (1 << i)) != 0) {
                    count++;
                }
            }
//            debug(bit, count);
            int index = bit % count;
            modArr[bit] = modArr[index] + 1;
        }

//        debug(modArr);

        for (int i = 0; i < n; i++) {
            int index;
            if (s.charAt(i) == '0') {
                index = (maMod + maArr[n - i - 1]) % (m + 1);
            } else {
                index = (mbMod - mbArr[n - i - 1] + (m - 1)) % (m - 1);
            }
            out.println(modArr[index] + 1);
        }

        // out.println("Hello World");
        out.flush();
    }

    void debug(Object... os) {
        System.err.println(Arrays.deepToString(os));
    }

    public static void main(String[] args) {
        new Main().run();
    }
}