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: Wrong Answer)

#[allow(unused_imports)]
use {
   proconio::{fastout, input, marker::*},
   std::cmp::*,
   std::collections::*,
};

#[fastout]
fn main() {
   input! {
      n: usize,
      x: Bytes,
   }
   let mut pc: usize = 0;
   for i in 0..n {
      if x[i] == b'1' {
         pc += 1;
      }
   }

   if pc == 0 {
      for _ in 0..n {
         println!("1");
      }
      return;
   }

   let mut m0: usize = 0;
   let mut m1: usize = 0;
   let mut r0: usize = 1;
   let mut r1: usize = 1;
   for i in (0..n).rev() {
      if x[i] == b'1' {
         m1 += r1 % (pc + 1);
      }
      r1 = (r1 * 2) % (pc + 1);
   }
   if pc > 1 {
      for i in (0..n).rev() {
         if x[i] == b'1' {
            m0 += r0 % (pc - 1);
         }
         r0 = (r0 * 2) % (pc - 1);
      }
   }
   let mut s = vec![0usize; pc + 1];
   let mut ans = vec![0; n];
   r0 = 1;
   r1 = 1;
   for i in (0..n).rev() {
      ans[i] = if x[i] == b'0' {
         f((m1 + r1) % (pc + 1), &mut s)
      } else {
         if pc == 1 {
            0
         } else {
            f((m0 + pc - 1 - r0) % (pc - 1), &mut s)
         }
      } + 1;
      if pc > 1 {
         r0 *= 2;
         r0 %= pc - 1;
      }
      r1 *= 2;
      r1 %= pc + 1;
   }
   for i in 0..n {
      println!("{}", ans[i]);
   }
}

fn f(m: usize, s: &mut Vec<usize>) -> usize {
   if m == 0 || s[m] != 0 {
      s[m]
   } else {
      let mut pc = 0;
      for i in 0.. {
         if (m >> i) & 1 == 1 {
            pc += 1;
         }
         if m >> i == 0 {
            break;
         }
      }
      let x = f(m % pc, s) + 1;
      s[m] = x;
      s[m]
   }
}