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.math.BigInteger;
import java.util.Arrays;
import java.util.Scanner;

class Main{
    static int maxn = 200005;
    static int[] f = new int[maxn];

    static int num(int a){
        int cnt = 0;
        while(a>0){
            if((a&1) > 0) ++cnt;
            a >>= 1;
        }
        return cnt;
    }
    static void pre(){
        f[0] = 0;
        f[1] = f[2] = 1;
        for(int i=3;i<maxn;++i){
            f[i] = 1+f[i%num(i)];
        }
    }
    static BigInteger kpow(BigInteger a,int b){
        BigInteger res = BigInteger.ONE;
        while(b>0){
            if((b&1) > 0) res = res.multiply(a);
            a = a.multiply(a);
            b >>= 1;
        }
        return res;
    }
    public static void main(String[] args){
        pre();

        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        String s = sc.next();
        sc.close();
        int cnt = 0;
        for(int i=0;i<n;++i)
            if(s.charAt(i)=='1') ++cnt;
        BigInteger dec = new BigInteger(s,2);
        for(int i=0;i<n;++i){
            int cnt1 = cnt;
            BigInteger d = dec;
            if(s.charAt(i)=='0'){
                ++cnt1;
                d = d.add(kpow(BigInteger.valueOf(2),n-i-1));
            }
            else{
                --cnt1;
                d = d.subtract(kpow(BigInteger.valueOf(2), n-i-1));
            }
            if(d.compareTo(BigInteger.valueOf(maxn))<0){
                int dd = d.intValue();
                System.out.println(f[dd]);
            }
            else{
                int dd = (d.mod(BigInteger.valueOf(cnt1))).intValue();
                System.out.println(1+f[dd]);
            }
        }
    }
}