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)

use proconio::{fastout, input};

use proconio::marker::Chars;
use std::cmp::max;

#[fastout]
fn main() {
    input! {
        n: usize,
        xx: Chars,
    }
    let mut m = vec![None; n + 10];
    fn f(x: usize, m: &mut [Option<usize>]) -> usize {
        if x == 0 {
            return 0;
        }
        if let Some(r) = m[x] {
            return r;
        } else {
            let r = 1 + f(x % x.count_ones() as usize, m);
            m[x] = Some(r);
            return r;
        }
    }
    let ones = xx.iter().filter(|&&x| x == '1').count() as u64;
    let mut a = 0;
    let mut c = 0;
    for i in 0..n {
        a *= 2;
        c *= 2;
        if xx[i] == '1' {
            a += 1;
            c += 1;
        }
        a %= max(ones, 2) - 1;
        c %= ones + 1;
    }
    let mut p = 1;
    let mut r = 1;
    let mut res = vec![0; n];
    for i in (0..n).rev() {
        p %= max(ones, 2) - 1;
        r %= ones + 1;
        if xx[i] == '0' {
            let x = (c + r) % (ones + 1);
            res[i] = 1 + f(x as usize, &mut m);
        }
        if xx[i] == '1' {
            if ones <= 1 {
                res[i] = 0;
            } else {
                let x = (ones - 1 + a - p) % (ones - 1);
                res[i] = 1 + f(x as usize, &mut m);
            }
        }
        p *= 2;
        r *= 2;
    }
    for i in 0..n {
        println!("{}", res[i]);
    }
}