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)

use std::io;
use std::io::prelude::*;

fn main() -> io::Result<()> {
    let n: usize = io::stdin().lock().lines().next().unwrap()?.parse().unwrap();
    let mut buf = vec![0; n];
    io::stdin().read(&mut buf)?;
    let cnt: u32 = buf.iter().filter(|&&b| b == b'1').count() as u32;
    let p0 = get_res_list(n, cnt + 1);
    if cnt == 1 {
        let lone_1 = buf.iter().position(|&b| b == b'1').unwrap();
        for i in 0..n {
            if i == lone_1 {
                println!("0");
            } else {
                let res = (p0[i] + p0[lone_1]) % (cnt + 1);
                println!("{}", 1 + f(res));
            }
        }
    } else {
        let p1 = get_res_list(n, cnt - 1);
        let mut res0 = 0; // 0の時使う
        let mut res1 = 0; // 1の時使う
        for (i, &b) in buf.iter().enumerate() {
            if b == b'1' {
                res0 += p0[i];
                res1 += p1[i];
            }
        }
        for (i, &bit) in buf.iter().enumerate() {
            let res = if bit == b'0' {
                (res0 + p0[i]) % (cnt + 1)
            } else {
                (res1 - p1[i]) % (cnt - 1)
            };
            println!("{}", 1 + f(res));
        }
    }
    Ok(())
}
/// 2^0, 2^1, ... をcntで割ったあまり
fn get_res_list(n: usize, cnt: u32) -> Vec<u32> {
    let mut v = vec![0; n];
    let mut res = 1;
    for i in 0..n {
        res %= cnt;
        if res == 0 {
            break;
        }
        v[n - i - 1] = res;
        res <<= 1;
    }
    v
}

fn f(mut x: u32) -> u32 {
    let mut cnt = 0;
    while x > 0 {
        x %= x.count_ones();
        cnt += 1;
    }
    cnt
}