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 1.8.0) (judge verdict: Runtime Error)


import java.util.*;

public class Main {
    static Scanner scanner;

    public static void main(String[] args) {
        scanner = new Scanner(System.in);
        int nS = nextInt();
        nextLine();
        String nextS = nextLine();
        for (int i = 0; i < nS; i++) {
            int r = 0;

            String next;
            StringBuilder sb = new StringBuilder(nextS);
            if (nextS.charAt(i) == '0') {
                sb.setCharAt(i,'1');
            } else {
                sb.setCharAt(i,'0');
            }
            next = sb.toString();
            int n = Integer.parseInt(next,2);
            while (n != 0) {
                n = n % count0(next);
                next = Integer.toBinaryString(n);
                r++;
            }
            System.out.println(r);
        }
    }

    public static int count0 (String a ) {
        char[] b = a.toCharArray();
        int r = 0;
        for (int i = 0; i < b.length; i++) {
            if (b[i] == '1') r++;
        }
        return r;
    }

    public static void println(Object o) {
        System.out.println(o);
    }

    public static void print(Object o) {
        System.out.print(o);
    }

    public static void printf(String format, Object... o) {
        System.out.printf(format, o);
    }

    public static int nextInt() {
        return Integer.parseInt(scanner.next());
    }

    public static long nextLong() {
        return Long.parseLong(scanner.next());
    }

    public static double nextDouble() {
        return Double.parseDouble(scanner.next());
    }

    public static short nextShort() {
        return Short.parseShort(scanner.next());
    }

    public static float nextFloat() {
        return Float.parseFloat(scanner.next());
    }

    public static void printYes() {
        System.out.println("Yes");
    }

    public static void printYES() {
        System.out.println("YES");
    }

    public static void printNo() {
        System.out.println("No");
    }

    public static void printNO() {
        System.out.println("NO");
    }

    public static String nextLine() {
        return scanner.nextLine();
    }

    public static double distance(double x, double y, double x2, double y2) {
        return Math.sqrt((x2 - x) * (x2 - x) + (y2 - y) * (y2 - y));
    }

    public static int MOD = 1_000_000_007;
    public static int MAX = 100000;
    public static long[] fac = new long[MAX];
    public static long[] finv = new long[MAX];
    public static long[] inv = new long[MAX];

    public static void combinit() {
        fac[0] = fac[1] = 1;
        finv[0] = finv[1] = 1;
        inv[1] = 1;
        for (int i = 2; i < MAX; i++) {
            fac[i] = fac[i - 1] * i % MOD;
            inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
            finv[i] = finv[i - 1] * inv[i] % MOD;
        }
    }

    public static long combination(int n, int k) {
        if (n < k || n < 0 || k < 0) return 0;
        return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
    }

    public static long factorial(long n) {
        return n <= 0 ? 1 : n * factorial(n - 1);
    }
}