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 Rust (1.42.0) (judge verdict: Runtime Error)

fn main() -> Result<(), Box<dyn std::error::Error>>{
  let mut n = String::new();
  let mut x = String::new();
  std::io::stdin().read_line(&mut n)?;
  std::io::stdin().read_line(&mut x)?;
  
  let n: usize = n.trim().parse()?;
  let x: Vec<u8> = x.trim().as_bytes().to_vec();
  let popcount: usize = x.iter().filter(|&b| *b == b'1').count();
  let mut dp = vec![0; n + 1];
  for i in 1 ..= n {
    let o = i.count_ones() as usize;
    dp[i] = dp[i % o] + 1;
  }

  let (mod1, score1, dp1) = prepare(n, popcount + 1, &x);
  let (mod2, score2, dp2) = prepare(n, popcount - 1, &x);
  
  for i in 1 ..= n {
    if x[i - 1] == b'0' {
      let z = (score1 + dp1[n - i]) % mod1;
      println!("{}", dp[z as usize] + 1);
    } else if popcount == 1 {
      println!("{}", 0);
    } else {
      let z = (score2 - dp2[n - i]) % mod2;
      println!("{}", dp[z as usize] + 1);
    }
  }
  Ok(())
}

fn prepare(n: usize, m: usize, x: &Vec<u8>) -> (i32, i32, Vec<i32>) {
  let m = m as i32;
  let mut score = 0;
  let mut dp = vec![1; n];

  if m == 0 {
    return (m, score, dp);
  }
  
  for i in 1 .. n {
    dp[i] = (dp[i - 1] * 2) % m;
  }
  for i in 1 ..= n {
    if x[i - 1] == b'1' {
      score += dp[n - i];
    }
  }
  (m, score, dp)
}