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: Accepted)



import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Scanner;

public final class Main {

    public static void main(String[] args) {
        final Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        final int n = Integer.parseInt(in.nextLine());
        final String num = in.nextLine();
        int bitCount = 0;
        for (int i = 0; i < n; i++) {
            bitCount += num.charAt(i) == '1' ? 1 : 0;
        }
        if (bitCount == 1) {
            for (int i = 0; i < n; i++) {
                if (num.charAt(i) == '1') {
                    System.out.println(0);
                } else {
                    if (i == n - 1 || num.charAt(n - 1) == '1') {
                        System.out.println(2);
                    } else {
                        System.out.println(1);
                    }
                }
            }
            return;
        }
        final long[] plusOne = new long[n];
        final long[] minusOne = new long[n];
        long tempPlusOne = 0;
        long tempMinusOne = 0;
        long runningPlus = 1;
        long runningMinus = 1;
        for (int i = n - 1; i >= 0; i--) {
            plusOne[i] = runningPlus;
            minusOne[i] = runningMinus;
            if (num.charAt(i) == '1') {
                tempPlusOne += runningPlus;
                tempMinusOne += runningMinus;
            }
            runningPlus = (runningPlus * 2) % (bitCount + 1);
            runningMinus = (runningMinus * 2) % (bitCount - 1);
        }
        for (int i = 0; i < n; i++) {
            if (num.charAt(i) == '0') {
                System.out.println(f((tempPlusOne + plusOne[i]) % (bitCount + 1)));
            } else {
                System.out.println(f((tempMinusOne - minusOne[i]) % (bitCount - 1)));
            }
        }
    }

    private static int f(long n) {
        int res = 1;
        while (n > 0) {
            n %= Long.bitCount(n);
            res++;
        }
        return res;
    }
}