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: Accepted)

#![allow(unused_imports)]
#![allow(non_snake_case, unused)]

use std::cmp::*;
use std::collections::*;
use std::ops::*;

// https://atcoder.jp/contests/hokudai-hitachi2019-1/submissions/10518254

macro_rules! eprint {
    ($($t:tt)*) => {{
        use ::std::io::Write;
        let _ = write!(::std::io::stderr(), $($t)*);
    }};
}
macro_rules! eprintln {
    () => { eprintln!(""); };
    ($($t:tt)*) => {{
        use ::std::io::Write;
        let _ = writeln!(::std::io::stderr(), $($t)*);
    }};
}
macro_rules! dbg {
    ($v:expr) => {{
        let val = $v;
        eprintln!("[{}:{}] {} = {:?}", file!(), line!(), stringify!($v), val);
        val
    }}
}

macro_rules! mat {
    ($($e:expr),*) => { Vec::from(vec![$($e),*]) };
    ($($e:expr,)*) => { Vec::from(vec![$($e),*]) };
    ($e:expr; $d:expr) => { Vec::from(vec![$e; $d]) };
    ($e:expr; $d:expr $(; $ds:expr)+) => { Vec::from(vec![mat![$e $(; $ds)*]; $d]) };
}

macro_rules! ok {
    ($a:ident$([$i:expr])*.$f:ident()$(@$t:ident)*) => {
        $a$([$i])*.$f($($t),*)
    };
    ($a:ident$([$i:expr])*.$f:ident($e:expr$(,$es:expr)*)$(@$t:ident)*) => { {
        let t = $e;
        ok!($a$([$i])*.$f($($es),*)$(@$t)*@t)
    } };
}

pub fn readln() -> String {
    let mut line = String::new();
    ::std::io::stdin().read_line(&mut line).unwrap_or_else(|e| panic!("{}", e));
    line
}

macro_rules! read {
    ($($t:tt),*; $n:expr) => {{
        let stdin = ::std::io::stdin();
        let ret = ::std::io::BufRead::lines(stdin.lock()).take($n).map(|line| {
            let line = line.unwrap();
            let mut it = line.split_whitespace();
            _read!(it; $($t),*)
        }).collect::<Vec<_>>();
        ret
    }};
    ($($t:tt),*) => {{
        let line = readln();
        let mut it = line.split_whitespace();
        _read!(it; $($t),*)
    }};
}

macro_rules! _read {
    ($it:ident; [char]) => {
        _read!($it; String).chars().collect::<Vec<_>>()
    };
    ($it:ident; [u8]) => {
        Vec::from(_read!($it; String).into_bytes())
    };
    ($it:ident; usize1) => {
        $it.next().unwrap_or_else(|| panic!("input mismatch")).parse::<usize>().unwrap_or_else(|e| panic!("{}", e)) - 1
    };
    ($it:ident; [usize1]) => {
        $it.map(|s| s.parse::<usize>().unwrap_or_else(|e| panic!("{}", e)) - 1).collect::<Vec<_>>()
    };
    ($it:ident; [$t:ty]) => {
        $it.map(|s| s.parse::<$t>().unwrap_or_else(|e| panic!("{}", e))).collect::<Vec<_>>()
    };
    ($it:ident; $t:ty) => {
        $it.next().unwrap_or_else(|| panic!("input mismatch")).parse::<$t>().unwrap_or_else(|e| panic!("{}", e))
    };
    ($it:ident; $($t:tt),+) => {
        ($(_read!($it; $t)),*)
    };
}


pub fn main() {
    let _ = ::std::thread::Builder::new().name("run".to_string()).stack_size(32 * 1024 * 1024).spawn(run).unwrap().join();
}

const MOD: usize = 1_000_000_007;
const INF: i32 = std::i32::MAX;

fn solve() {
    let n = read!(usize);
    let c = read!([char]);
    let mut x = 0;
    for i in 0..n {
        if c[i]=='1' {
            x += 1;
        }
    }
    let mut dp = vec![vec![0;2];n];
    let mut val = vec![0;n];
    let mut base = 1;
    for i in 0..n {
        let res = base % (x+1);
        dp[i][0] = res;
        base *= 2;
        base %= (x+1);
        if c[n-1-i]=='1'{
            val[0] += res;
            val[0] %= (x+1);
        }
    }
    if x>=2 {
        base = 1;
        for i in 0..n {
            let res = base % (x-1);
            dp[i][1] = res;
            base *= 2;
            base %= (x-1);
            if c[n-1-i]=='1'{
                val[1] += res;
                val[1] %= (x-1);
            }
        }
    }
    // if x==0 {
    //     println!("0");
    //     return;
    // }
    let mut ans = vec![0;n+1];
    for i in 1..n+1 {
        let mut sum = 0;
        let mut base = i;
        loop {
            let mut d = base.count_ones();
            let rem = base % (d as usize);
            base = rem;
            sum += 1;
            if base==0 {
                break;
            }
        }
        ans[i] = sum;
    }
    if x<=1 {
        for i in 0..n {
            let mut sum = 0;
            if c[i]=='0' {
                let mut base = val[0];
                base += dp[n-1-i][0];
                base %= (x+1);
                sum += 1;
                sum += ans[base];
                println!("{}",sum);
            }
            if c[i]=='1' {
                println!("0");
                continue;
            }
        }
        return;
    }
    for i in 0..n {
        let mut sum = 0;
        if c[i]=='0' {
            let mut base = val[0];
            base += dp[n-1-i][0];
            base %= (x+1);
            sum += 1;
            sum += ans[base];
            println!("{}",sum);
        }
        if c[i]=='1' {
            let mut base = val[1];
            base += x-1;
            base -= dp[n-1-i][1];
            base %= (x-1);
            sum += 1;
            sum += ans[base];
            println!("{}",sum);
        }
    }
}

fn run() {
    solve();
}