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

public class Main {
    private HashMap<Long, HashMap<Integer, Long>> pow_cache = new HashMap<>();
    public static void main(String[] args) {
        Main main = new Main();
        main.solve();
    }
    public void solve() {
        Scanner scan = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int N = scan.nextInt();
        String X = scan.next();
        char[] data = X.toCharArray();
        HashMap<Long, Long> cache = new HashMap<>();
        cache.put(0L, 1L);
        for (int i = 0; i < N; i++) {
            int count = 0;
            for (int j = 0; j < N; j++) {
                if (i == j) {
                    count += data[j] == '0' ? 1 : 0;
                    continue;
                }
                if (data[j] == '1') {
                    count += 1;
                }
            }
            long current = 0;
            for (int j = 0; j < N; j++) {
                if (i == j && data[j] == '1') {
                    continue;
                }
                if (i != j && data[j] == '0') {
                    continue;
                }
                current += pow(2, N - 1 - j, count);
                current %= count;
            }
            long ans = 1;
            if (cache.containsKey(current)) {
                System.out.println(cache.get(current));
                continue;
            }
            long previous = current;
            while (current != 0) {
                char[] x = Long.toBinaryString(current).toCharArray();
                int len = x.length;
                int c = 0;
                for (int j = 0; j < len; j++) {
                    if (x[j] == '1') {
                        c += 1;
                    }
                }
                current = 0;
                for (int j = 0; j < len; j++) {
                    if (x[j] == '0') {
                        continue;
                    }
                    current += pow(2, len - 1 - j, c);
                    current %= c;
                }
                ans += 1;
            }
            cache.put(previous, ans);
            System.out.println(ans);
        }
    }
    private long pow(int base, int count, long mod) {
        if (!pow_cache.containsKey(mod)) {
            pow_cache.put(mod, new HashMap<>());
        }
        HashMap<Integer, Long> cache = pow_cache.get(mod);
        if (cache.containsKey(count)) {
            return cache.get(count);
        }
        if (count == 0) {
            long t = 1 % mod;
            cache.put(0, t);
            return t;
        }
        if (count % 2 == 1) {
            long t = base * pow(base, count - 1, mod);
            t %= mod;
            cache.put(count, t);
            return t;
        }
        long t = pow(base, count / 2, mod);
        t *= t;
        t %= mod;
        cache.put(count, t);
        return t;
    }
}