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)

#![allow(unused, non_snake_case, dead_code, non_upper_case_globals)]
use {
    proconio::{marker::*, *},
    std::{cmp::*, collections::*, mem::*},
};

#[fastout]
fn main() {
    input! {//
        n:usize,
        mut x:Chars,
    }
    x.reverse();
    let mut basecount = 0i64;

    // めちゃめちゃすくない

    for &c in x.iter() {
        if c == '1' {
            basecount += 1;
        }
    }
    // dbg!(basecount);
    // let mut dp = vec![vec![0; 2]; n + 1];
    let mut dp = vec![0; 201010];
    for i in 0..201010usize {
        if i == 0 {
            dp[0] = 0;
        }
        if i == 1 {
            dp[1] = 1;
        } else {
            let mut tmp = i;
            let mut cnt = 0;
            while tmp > 0 {
                // dbg!(tmp, tmp.count_ones());
                tmp %= tmp.count_ones() as usize;
                cnt += 1;
            }
            dp[i] = cnt;
        }
    }
    // dbg!(&dp);

    let mut fbasep = 0;
    let mut fbasen = 0;
    for i in 0..n {
        if x[i] == '1' {
            fbasep += mod_pow(2, i as i64, basecount + 1);
            fbasen += mod_pow(2, i as i64, basecount - 1);
        }
        // fbase %= basecount;
    }
    // println!("{}", fbase);

    for i in (0..n).rev() {
        let mut bc = basecount;
        let mut pn = 1i64;
        if x[i] == '1' {
            pn = -1;
        } else {
            pn = 1;
        }
        bc += pn;
        if pn == 1 {
            let mut b = fbasep + mod_pow(2, i as i64, bc);
            b %= bc;
            println!("{}", dp[b as usize] + 1);
        } else {
            let mut b = fbasen + bc - mod_pow(2, i as i64, bc);
            b %= bc;
            println!("{}", dp[b as usize] + 1);
        }
        // if x[i] == '1' {
        //     bc += 1;
        // } else {
        //     bc -= 1;
        // }
        // let mut b = fbase + if x[i] == '1' { 1 } else { -1 } * mod_pow(2, i as i64, MX);
        // b = (b + bc * 2) % bc;
        // // dbg!(i, b);
        // println!("{}", dp[b as usize] + 1);
    }
}
const MX: i64 = 200100;
const INF: i64 = 1 << 60;
/// x ^ n % m
pub fn mod_pow(x: i64, n: i64, m: i64) -> i64 {
    let mut res = 1;
    let mut x = x % m;
    let mut n = n;
    while n > 0 {
        if n & 1 == 1 {
            res = (res * x) % m;
        }
        x = (x * x) % m;
        n >>= 1;
    }
    res
}