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.util.*;

public class Main {
    static int MOD = 1000000009;
    
    public static void main(String[] args) throws Exception {
        // Your code here!
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        String str = sc.next();
        char[] c = str.toCharArray();
        
        int originalPopcount = 0;
        for (int i = 0; i < n; i++) {
            if (c[i] == '1') originalPopcount++;
        }
        
        if (originalPopcount == 0) {
            for (int i = 0; i < n; i++) {
                System.out.println(0);
            }
            return;
        }
        
        int plusPopcount = originalPopcount + 1;
        int minusPopcount = originalPopcount - 1;
        
        long[] plusMod = new long[n+1];
        int digit = 1;
        int m = plusPopcount;
        for (int i = n - 1; i >= 0; i--) {
            long num = (c[i] - '0') * digit;
            num %= m;
            
            plusMod[i] = plusMod[i+1] + num;
            plusMod[i] %= m;
            
            digit *= 2;
            digit %= m;
        }
        
        long[] minusMod = new long[n+1];
        digit = 1;
        m = minusPopcount;
        if (m != 0) {
            for (int i = n - 1; i >= 0; i--) {
                long num = (c[i] - '0') * digit;
                num %= m;
                
                minusMod[i] = minusMod[i+1] + num;
                minusMod[i] %= m;
                
                digit *= 2;
                digit %= m;
            }
        }
        
        for (int i = 0; i < n; i++) {
            int count = 0;
            
            int f = 0;
            if (c[i] == '0') {
                f = (int)(plusMod[0] + modpow(2, n-1-i, plusPopcount));
                f %= plusPopcount;
                count++;
            } else {
                if (minusPopcount == 0) {
                    System.out.println(0);
                    continue;
                }
                
                f = (int)(minusMod[0] - modpow(2, n-1-i, minusPopcount));
                f %= minusPopcount;
                count++;
                if (f < 0) {
                    f += minusPopcount;
                }
            }
            // System.out.println(f);
            
            while (f > 0) {
                int popcount = Integer.bitCount(f);
                f %= popcount;
                count++;
            }
            System.out.println(count);
        }
    }
    
    public static long modpow(long num, long n, int mod) {
        // ex. 3^10
        // 3^10 = 3^(0b1010)
        // = 3^8が1個 * 3^4が0個 * 3^2が1個 * 3^1が0個
        // (次の桁の値は(前の桁)^2になる)
        
        long res = 1;
        long digit = num;
        
        while (n > 0) {
            long lsb = n & 1;
            if (lsb == 1) {
                res *= digit;
                res %= mod;
            }
            
            digit = digit * digit;
            digit %= mod;
            n = n >> 1;
        }
        
        return res;
    }
}