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)

// https://atcoder.jp/contests/aising2020/tasks/aising2020_d

use proconio::fastout;
use proconio::input;

#[fastout]
fn main() {
    input! {
       n: u32,
       x: String,
    }
    let x: Vec<char> = x.chars().collect();
    let x = to_u32(x, n);
    for i in 1..=n {
        let ans = f(xi(x, n - i));
        println!("{}", ans);
    }
}

fn to_u32(xs: Vec<char>, n: u32) -> u32 {
    let mut ans = 0;
    for (i, &x) in xs.iter().enumerate() {
        if x == '1' {
            ans += 2u32.pow(n - 1 - i as u32);
        }
    }
    ans
}

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

fn xi(x: u32, j: u32) -> u32 {
    x ^ 2u32.pow(j)
}