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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;

public class Main {

  static int n;
  static long ans;
  static int[] dp;
  static int[] op;
  static String x;

  public static void main(String[] args) throws Exception {

    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter out = new PrintWriter(System.out);

    String[] in = br.readLine().split(" ");
    n = Integer.parseInt(in[0]);

    x = br.readLine();

    br.close();

    op = new int[n+1];

    dp = new int[200001];

    dp[0] = 0;

    for (int i = 1; i <= 200000; i++) {
      dp[i] = dp[i % Integer.bitCount(i)] + 1;
    }

    int initBit = 0;

    for (int i = 0; i < n; i++) {
      if (x.charAt(i) == '1') {
        initBit++;
      }
    }

    int u = 0, l = 0;

    for (int i = n-1; i >= 0; i--) {
      u = 2 * u + (x.charAt(i) - '0');
      l = 2 * l + (x.charAt(i) - '0');
      u %= initBit+1;
      if(initBit > 1){
        l %= initBit-1;
      }
    }

    int adu = 1%(initBit+1), adl = 0;
    if (initBit > 1) adl = 1 % (initBit-1);

    for (int i = n-1; i >= 0; i--) {

      if (x.charAt(i) == '0') {
        op[i] = dp[(u + adu) % (initBit + 1)] + 1;
      } else {
        if (initBit > 1) {
          op[i] = dp[(l - adl + initBit - 1) % (initBit - 1)] + 1;
        } else {
          op[i] = 0;
        }
      }

      adu = (adu * 2) % (initBit + 1);
      if(initBit > 1) adl = (adl * 2) % (initBit - 1);

    }

    for (int i = 0; i < n; i++) {
      out.println(op[i]);
    }

    out.flush();
  }
}