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.PrintWriter;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		char[] x = sc.next().toCharArray();
		sc.close();

		int cnt = 0;
		for (int i = 0; i < n; i++) {
			if (x[i] == '1') {
				cnt++;
			}
		}
		int p0 = cnt + 1;
		int p1 = cnt - 1;
		int[] dp = new int[n];
		for (int i = 1; i < n; i++) {
			int next = i % Integer.bitCount(i);
			dp[i] = dp[next] + 1;
		}

		long v0 = 0;
		for (int i = 0; i < n; i++) {
			v0 *= 2;
			v0 %= p0;
			if (x[i] == '1') {
				v0++;
			}
		}
		v0 %= p0;

		long v1 = 0;
		if (p1 > 0) {
			for (int i = 0; i < n; i++) {
				v1 *= 2;
				v1 %= p1;
				if (x[i] == '1') {
					v1++;
				}
			}
			v1 %= p1;
		}

		PrintWriter pw = new PrintWriter(System.out);
		for (int i = 0; i < n; i++) {
			if (x[i] == '0') {
				long a = power(2, n - 1 - i, p0);
				long val = v0 + a;
				int next = (int) (val % p0);
				pw.println(dp[next] + 1);
			} else {
				if (v1 == 0) {
					pw.println(0);
				} else {
					long a = power(2, n - 1 - i, p1);
					long val = v1 - a + p1;
					int next = (int) (val % p1);
					pw.println(dp[next] + 1);
				}
			}
		}
		pw.flush();
	}

	static long power(long x, long n, int m) {
		if (n == 0) {
			return 1;
		}
		long val = power(x, n / 2, m);
		val = val * val % m;
		if (n % 2 == 1) {
			val = val * x % m;
		}
		return val;
	}
}