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)

#[allow(unused_imports)]
use proconio::{
    fastout, input,
    marker::{Bytes, Chars, Isize1, Usize1},
};

fn pow(mut s: usize, mut n: usize, m: usize) -> usize {
    if m == 0 {
        return 1;
    }
    let mut temp = s;
    s = 1;
    while n > 0 {
        if n & 1 != 0 {
            s *= temp;
            s %= m;
        }
        n >>= 1;
        temp *= temp;
        temp %= m;
    }
    s
}

fn func(n: usize) -> usize {
    if n == 0 {
        return 0;
    }
    1 + func(n % n.count_ones() as usize)
}

#[fastout]
fn main() {
    input! {
        n:usize,
        x:Chars
    }
    let one_count = x.iter().filter(|&&z| z == '1').count();

    if one_count == 0 {
        for _ in 0..n {
            println!("{}", 1);
        }
        return;
    }

    //let mut table = vec![0_usize; n + 1];
    let mut tablep1 = vec![0_usize; n + 1];
    let mut tablem1 = vec![0_usize; n + 1];
    for i in 0..n {
        if x[i] == '1' {
            //table[i + 1] = pow(2, n - i - 1, one_count);
            tablep1[i + 1] = pow(2, n - i - 1, one_count + 1);
            tablem1[i + 1] = pow(2, n - i - 1, one_count - 1);
        }
    }

    for i in 1..=n {
        //table[i] = (table[i - 1] + table[i]) % one_count;
        tablep1[i] = (tablep1[i - 1] + tablep1[i]) % (one_count + 1);
        tablem1[i] = if one_count == 1 {
            0
        } else {
            (tablem1[i - 1] + tablem1[i]) % (one_count - 1)
        };
    }

    for i in 0..n {
        if x[i] == '1' {
            if one_count == 1 {
                println!("{}", 0);
            } else {
                let buf =
                    (tablem1[i] + tablem1[n] + (one_count - 1) - tablem1[i + 1]) % (one_count - 1);
                println!("{}", 1 + func(buf));
            }
        } else {
            let buf = (tablep1[i] + tablep1[n] + (one_count + 1) - tablep1[i + 1]
                + pow(2, n - i - 1, one_count + 1))
                % (one_count + 1);
            println!("{}", 1 + func(buf));
        }
    }
}