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 itertools::Itertools;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;

fn main() {
    input! {
        n: usize,
        x: Chars,
    }

    let p = x.iter().filter(|&c| *c == '1').count() as i64;

    let mut p2p1 = 0;
    let mut d = 1 % (p + 1);
    let mut dp1 = vec![0; n];
    for i in (0..n).rev() {
        if x[i] == '1' {
            p2p1 += d;
            p2p1 %= p + 1;
        }
        dp1[i] = d;
        d *= 2;
        d %= p + 1;
    }
    let mut p2m1 = 0;
    let mut dm1 = vec![0; n];
    if p > 1 {
        let mut d = 1 % (p - 1);
        for i in (0..n).rev() {
            if x[i] == '1' {
                p2m1 += d;
                p2m1 %= p - 1;
            }
            dm1[i] = d;
            d *= 2;
            d %= p - 1;
        }
    }

    for i in 0..n {
        let p2;
        if x[i] == '1' {
            p2 = p - 1;
        } else {
            p2 = p + 1;
        }
        if p2 == 0 {
            println!("0");
            continue;
        }
        if p2 == 1 {
            println!("1");
            continue;
        }

        let mut y;
        if x[i] == '0' {
            y = p2p1;
            y += dp1[i];
            y %= p2;
        } else {
            y = p2m1;
            y += p2 - dm1[i];
            y %= p2;
        }

        let mut res = 1;
        while y != 0 {
            y %= y.count_ones() as i64;
            res += 1;
        }
        println!("{}", res);
    }
}