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)

use proconio::{fastout, input};

#[fastout]
fn main() {
    input! {
        n: usize,
        x: String,
    }
    let x: i32 = parse(&x);

    for i in (0..n).rev() {
        let mut x = x ^ 1 << i;
        for i in 1.. {
            let count_one = count_one(x);
            x %= count_one;
            if x == 0 {
                println!("{}", i);
                break;
            }
        }
    }
}

fn parse(s: &str) -> i32 {
    let mut res = 0;
    for (i, c) in s.chars().rev().enumerate() {
        match c {
            '1' => res += 2i32.pow(i as u32),
            _ => continue,
        }
    }
    res
}

fn count_one(n: i32) -> i32 {
    let mut res = 0;
    for i in 0..32 {
        if n & 1 << i != 0 {
            res += 1;
        }
    }
    res
}