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)

// -*- coding:utf-8-unix -*-

use proconio::input;

fn popcount(n: u64) -> u64 {
    let mut num_of_one = 0;
    for i in 0..n {
        let shifted = n >> i;
        if shifted == 0 {
            break;
        }

        if shifted & 1 == 1 {
            num_of_one += 1;
        }
    }
    num_of_one
}

#[test]
fn test_popcount() {
    assert_eq!(popcount(3), 2);
    assert_eq!(popcount(7), 3);
    assert_eq!(popcount(0), 0);
}

fn main() {
    input! {
        n: u64,
        s: String,
    }

    let x: Vec<char> = s.chars().collect();
    let mut ans = Vec::with_capacity((n + 1) as usize);
    for _i in 0..n + 1 {
        ans.push(-1);
    }

    for i in 1..n + 1 {
        if ans[i as usize] >= 0 {
            continue;
        }

        let mut x_now = x.clone();
        if x_now[(i - 1) as usize] == '0' {
            x_now[(i - 1) as usize] = '1';
        } else {
            x_now[(i - 1) as usize] = '0';
        }
        if x_now.iter().all(|c| *c == '0') {
            ans[i as usize] = 0;
            continue;
        }

        let mut calc_num = 0;
        while !x_now.iter().all(|c| *c == '0') {
            let x_num = u64::from_str_radix(&x_now.iter().collect::<String>(), 2).unwrap();
            let x_num_after = x_num % popcount(x_num);
            calc_num += 1;
            // if ans[x_num_after as usize] >= 0 {
            //     ans[i as usize] = calc_num + ans[x_num_after as usize];
            //     break;
            // } else if x_num_after == 0 {
            //     ans[i as usize] = calc_num;
            //     break;
            // }
        if x_num_after == 0 {
            ans[i as usize] = calc_num;
            break;
        }


            // 桁消える
            let s_tmp = format!("{:b}", x_num_after);
            x_now = s_tmp.chars().collect();
        }
    }

    for i in 1..n + 1 {
        println!("{}", ans[i as usize]);
    }
}