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)

use std::cmp::*;
use std::io::*;

fn calc(mut n: i64, mut x: i64) -> i64 {
    let mut res = 0;
    while n != 0 {
        if x == 0 {
            return 0;
        }
        n = n % x;
        x = n.count_ones() as i64;
        res += 1;
    }
    res
}

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: i64 = itr.next().unwrap().parse().unwrap();
    let mut s: Vec<char> = itr.next().unwrap().chars().collect();

    let mut cnt = 0;
    for i in 0..n {
        if s[i as usize] == '1' {
            cnt += 1;
        }
    }
    s.reverse();

    let mut pc1 = vec![1; n as usize + 1];
    let mut pc2 = vec![1; n as usize + 1];
    for i in 0..n {
        pc1[i as usize + 1] = pc1[i as usize] * 2 % (cnt + 1);
        if cnt > 1 {
            pc2[i as usize + 1] = pc2[i as usize] * 2 % (cnt - 1);
        }
    }

    let mut all1 = 0;
    let mut all2 = 0;
    for i in 0..n {
        all1 += if s[i as usize] == '1' {
            pc1[i as usize]
        } else {
            0
        };
        all1 %= cnt + 1;

        if cnt > 1 {
            all2 += if s[i as usize] == '1' {
                pc2[i as usize]
            } else {
                0
            };
            all2 %= cnt - 1;
        }
    }

    let mut out = Vec::new();
    for i in 0..n {
        if s[i as usize] == '1' {
            if cnt <= 1 {
                writeln!(out, "{}", 0).ok();
            } else {
                let tmp = (cnt - 1 + all2 - pc2[i as usize] % (cnt - 1)) % (cnt - 1);
                writeln!(out, "{}", calc(tmp, tmp.count_ones() as i64) + 1).ok();
            }
        } else {
            let mut tmp = all1 + pc1[i as usize];
            tmp %= cnt + 1;
            writeln!(out, "{}", calc(tmp, tmp.count_ones() as i64) + 1).ok();
        }
    }

    out.reverse();
    stdout().write_all(&out).unwrap();
}