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: Accepted)


import java.util.*;

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    String x = sc.next();
    long p = 0;
    for(int i = 0; i < n; i++) {
      if(x.charAt(i) == '1') p++;
    }
    if(p == 0) {
      for(int i = 0; i < n; i++) {
        System.out.println(1); 
      }
    } else {
      long[] r0 = new long[n];
      r0[n - 1] = 1;
      for(int i = n - 2; i >= 0; i--) {
        r0[i] = (2 * r0[i + 1]) % p;
      }
      long t0 = 0;
      for(int i = 0; i < n; i++) {
        if(x.charAt(i) == '1') t0 = (t0 + r0[i]) % p;
      }

      long[] r1 = new long[n];
      r1[n - 1] = 1;
      for(int i = n - 2; i >= 0; i--) {
        r1[i] = (2 * r1[i + 1]) % (p + 1);
      }
      long t1 = 0;
      for(int i = 0; i < n; i++) {
        if(x.charAt(i) == '1') t1 = (t1 + r1[i]) % (p + 1);
      }

      long[] r2 = new long[n];
      long t2 = 0;
      r2[n - 1] = 1;
      if(p > 1) {
        for(int i = n - 2; i >= 0; i--) {
          r2[i] = (2 * r2[i + 1]) % (p - 1);
        }
        for(int i = 0; i < n; i++) {
          if(x.charAt(i) == '1') t2 = (t2 + r2[i]) % (p - 1);
        }
      }

      for(int i = 0; i < n; i++) {
        if(x.charAt(i) == '0') {
          long s = (t1 + r1[i]) % (p + 1);
          int ans = 1;
          while(s > 0) {
            ans++;
            int d = 0;
            long s1 = s;
            while(s1 > 0) {
              d += (s1 % 2);
              s1 /= 2;
            }
            s = (s % d);
          }
          System.out.println(ans);
        } else {
          if(p == 1) {
            System.out.println(0); 
          } else {
            long s = (t2 - r2[i] + p - 1) % (p - 1);
            int ans = 1;
            while(s > 0) {
              ans++;
              int d = 0;
              long s1 = s;
              while(s1 > 0) {
                d += (s1 % 2);
                s1 /= 2;
              }
              s = (s % d);
            }
            System.out.println(ans);
          }
        }
      }
    }
  }
}