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: Time Limit Exceeded)

use num_bigint::{BigInt, ToBigInt};
use num_traits::Zero;
use proconio::marker::Bytes;
use proconio::{fastout, input};

fn g(x: usize) -> usize {
    if x == 0 {
        return 0;
    }
    g(x % (x as u64).count_ones() as usize) + 1
}
fn f(x: &BigInt, pops: u64) -> usize {
    if x.is_zero() {
        return 0;
    }
    let tmp = x % pops.to_bigint().unwrap();
    let tmp: usize = tmp.to_str_radix(10).parse().unwrap();
    g(tmp) + 1
}

#[fastout]
fn main() {
    input! {
        n: usize,
        mut s: Bytes,
    }

    s.reverse();
    let mut t: BigInt = 0.to_bigint().unwrap();
    let mut k: BigInt = 1.to_bigint().unwrap();
    let mut pop_counts: usize = 0;
    for i in 0..n {
        if s[i] == '1' as u8 {
            t += &k;
            pop_counts += 1;
        }
        k *= 2.to_bigint().unwrap();
    }

    let mut k: BigInt = 1.to_bigint().unwrap();
    let mut ans = vec![0; n];
    for i in 0..n {
        if s[i] == '1' as u8 {
            ans[i] = f(&(&t - &k), (pop_counts - 1) as u64);
        } else {
            ans[i] = f(&(&t + &k), (pop_counts + 1) as u64);
        }
        k *= 2.to_bigint().unwrap();
    }
    ans.reverse();
    for i in 0..n {
        println!("{}", ans[i]);
    }
}