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::input;

#[allow(unused_variables)]
fn main() {
    input! {
      n: usize,
      x: String
    };

    // let mut ans = Vec::new();

    for i in 0..n {
        let reversed_x = reverse_bit(x.clone(), i);
        let mut ans = 0;
        // 反転後の10進数の値
        let mut a = usize::from_str_radix(&reversed_x, 2).unwrap();

        let mut flag = false;

        loop {
            if flag {
                a = a % to_bin(a).chars().filter(|&v| v == '1').count();
            } else {
                a = a % &reversed_x.chars().filter(|&v| v == '1').count();
                flag = true;
            }
            ans += 1;

            if a == 0 {
                break;
            }
        }

        println!("{}", ans);
    }
}

fn to_bin(d: usize) -> String {
    let mut decimal = d;
    let mut s = String::new();

    while decimal > 0 {
        let m = decimal % 2;
        decimal = decimal / 2;
        s = format!("{}{}", s, m);
    }

    s
}

fn reverse_bit(v: String, i: usize) -> String {
    let mut ca: Vec<char> = v.chars().collect();
    if ca[i] == '0' {
        ca[i] = '1'
    } else {
        ca[i] = '0'
    }

    ca.into_iter().collect()
}