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 1.8.0) (judge verdict: Time Limit Exceeded)

import java.math.BigInteger;
import java.util.*;
import java.util.stream.Collectors;

public class Main {
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    init();
    int n = sc.nextInt();
    String x = sc.next();
    char[] xc = x.toCharArray();
    int baseCount = 0;
    for (int j = 0, len = x.length(); j < len; j++) {
      if (xc[j] == '1') {
        baseCount++;
      }
    }
    BigInteger xB = new BigInteger(x, 2);
    BigInteger count1 = BigInteger.valueOf(baseCount - 1);
    BigInteger count2 = BigInteger.valueOf(baseCount + 1);
    for (int i = 1; i <= n; i++) {
      char tmp = xc[i - 1];
      BigInteger t = tmp == '1' ? count1 : count2;
      BigInteger newB;
      if (tmp == '1') {
        newB = xB.clearBit(n - i);
      } else {
        newB = xB.setBit(n - i);
      }
      System.out.println(newB.compareTo(BigInteger.ZERO) == 0 ? 0 : map.get(newB.remainder(t).intValue()) + 1);
    }
  }

  private static Map<Integer, Integer> map = new HashMap<>();
  private static void init() {
    int limit = 200000;
    map.put(0, 0);
    map.put(1, 1);
    for (int i = 2; i <= limit; i++) {
      String b = Integer.toString(i, 2);
      int count = 0;
      for (int j = 0, len = b.length(); j < len; j++) {
        if (b.charAt(j) == '1') {
          count++;
        }
      }
      int t = map.get(i % count);
      map.put(i, t + 1);
    }
  }
}