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 {

  private static void solve(String X, int N) {
    char[] A = X.toCharArray();
    int bitCnt = 0;
    for (char c : A) {
      if (c == '1') {
        bitCnt++;
      }
    }
    if (bitCnt == 0) {
      for (int i = 0; i < N; i++) {
        System.out.println(1);
      }
    } else {
      int[] pow1 = new int[N];
      int[] pow0 = new int[N];
      pow1[0] = 1;
      pow0[0] = 1;
      for (int i = 1; i < N; i++) {
        pow1[i] = (2 * pow1[i - 1]) % (bitCnt + 1);
        if (bitCnt == 1) continue;
        pow0[i] = (2 * pow0[i - 1]) % (bitCnt - 1);
      }
      int sum1 = 0;
      int sum0 = 0;
      for (int i = 0; i < N; i++) {
        if (A[i] == '1') {
          sum1 = (sum1 + pow1[N - 1 - i]) % (bitCnt + 1);
          if (bitCnt == 1) continue;
          sum0 = (sum0 + pow0[N - 1 - i]) % (bitCnt - 1);
        }
      }
      for (int i = 0; i < N; i++) {
        if (A[i] == '1') {
          if (bitCnt == 1) {
            System.out.println(0); // val % 1 = 0
            continue;
          }
          int val = (sum0 - pow0[N - 1 - i] + (bitCnt - 1)) % (bitCnt - 1);
          System.out.println(getStepsCnt(val));
        } else {
          int val = (sum1 + pow1[N - 1 - i]) % (bitCnt + 1);
          System.out.println(getStepsCnt(val));
        }
      }
    }
  }

  private static int getStepsCnt(int val) {
    int cnt = 1;
    while (val > 0) {
      val = val % Integer.bitCount(val);
      cnt++;
    }
    return cnt;
  }

  public static void main(String[] args) {
    int N = scanner.nextInt();
    String X = scanner.next();
    solve(X, N);
  }

  private static final Scanner scanner = new Scanner(System.in);
}