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: Time Limit Exceeded)

use itertools::Itertools;
use proconio::input;
use proconio::marker::{Chars, Usize1};
use std::cmp::*;
use std::collections::*;
use std::iter::Iterator;

#[allow(unused_macros)]
macro_rules! max {
    ($x:expr) => {
      $x
    };
    ($x:expr, $($xs:tt)+) => {
      max($x,max!($($xs)+))
    };
  }

#[allow(unused_macros)]
macro_rules! min {
    ($x:expr) => {
      $x
    };
    ($x:expr, $($xs:tt)+) => {
      min($x,min!($($xs)+))
    };
  }

#[allow(unused_macros)]
macro_rules! debug {
    ($($a:expr),*) => {
        eprintln!(concat!($(stringify!($a), " = {:?}, "),*), $($a),*);
    }
}

fn main() {
    input! {
    n: u64,
    x: Chars
    }
    
    let mut num = 0;
    for i in 0..n {
        if x[i as usize] == '1' {
            num += 2u64.pow((n-i-1) as u32);
        }
    }

    for i in 0..n {
        let mask = 2u64.pow((n-i-1) as u32);
        let target = num ^ mask;
        println!("{}", f(target, n as usize));
    }
}

fn f(x: u64, keta : usize) -> u64 {
    let mut cnt = 0;

    let mut x = x;
    while x != 0 {
        let s = format!("{:0>1$b}", x, keta);
        let mut onecnt = 0;
        for op in 0..keta {
            if x >> op & 1 == 1 {
                onecnt += 1;
            }
        }

        x = x % onecnt;
        cnt += 1;
    }

    return cnt;
}