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 proconio::*;

fn main() {
    input! {
        n: usize,
        s: String
    }

    let mut ans = vec![0; n];
    let x: Vec<i64> = s.chars().map(|e| e.to_digit(2).unwrap() as i64).collect();
    let pc: i64 = x.iter().sum();
    for b in 0..=1 {
        let mut pc = pc;
        if b == 0 {
            pc += 1;
        } else {
            if pc == 1 {
                continue;
            }
            pc -= 1;
        }

        let mut r0: i64 = 0;
        for &e in &x {
            r0 = ((r0 << 1) + e).rem_euclid(pc);
        }

        let mut k: i64 = 1;
        for (i, &e) in x.iter().enumerate().rev() {
            if e == b {
                if b == 0 {
                    ans[i] = f((r0 + k).rem_euclid(pc)) + 1;
                } else {
                    ans[i] = f((r0 - k).rem_euclid(pc)) + 1;
                }
            }
            k = (k << 1).rem_euclid(pc);
        }
    }

    for e in &ans {
        println!("{}", e);
    }
}

fn f(n: i64) -> i64 {
    if n == 0 {
        return 0;
    }

    1 + f(n.rem_euclid(n.count_ones() as i64))
}