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.Scanner;

public class Main {

    public int binaryToDecimal(String binary) {
        return Integer.parseInt(binary, 2);
    }

    public int countOnes(String binary) {
        int count = 0;
        for (char d : binary.toCharArray()) {
            if (d == '1') {
                count++;
            }
        }
        return count;
    }

    public String decimalToBinary(int decimal) {
        return Integer.toBinaryString(decimal);
    }

    public int modulus(String str, int n, int k) {
        int pT[] = new int[n];
        pT[0] = 1 % k;
        for (int i = 1; i < n; i++) {
            pT[i] = pT[i - 1] * (2 % k);
            pT[i] %= k;
        }
        int res = 0;
        int i = 0, j = n - 1;
        while (i < n) {
            if (str.charAt(j) == '1') {
                res += (pT[i]);
                res %= k;
            }
            i++;
            j--;
        }
        return res;
    }

    public static void main(String args[]) {
        Main azo = new Main();
        Scanner sc = new Scanner(System.in);
        StringBuilder sb = new StringBuilder();
        int n = sc.nextInt();
        String x = sc.next();
        char[] xarr = x.toCharArray();
        for (int i = 0; i < xarr.length; i++) {
            int count = 0;
            if (xarr[i] == '1') {
                xarr[i] = '0';
            } else {
                xarr[i] = '1';
            }
            String binaryString = String.valueOf(xarr);
            while (azo.countOnes(binaryString) != 0) {
                count++;
                int ones = azo.countOnes(binaryString);
                int val = azo.modulus(binaryString, binaryString.length(), ones);
                binaryString = azo.decimalToBinary(val);
            }
            if (xarr[i] == '1') {
                xarr[i] = '0';
            } else {
                xarr[i] = '1';
            }
            sb.append(count).append("\n");
        }
        System.out.println(sb.toString());
    }
}