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 std::cmp::*;
use std::collections::*;
use itertools::Itertools;
use num_traits::clamp;
use ordered_float::OrderedFloat;
use proconio::{input, marker::*, fastout};
use superslice::*;

#[fastout()]
fn main() {
    input! {
        n: usize,
        x: Chars,
    }
    // let x = [1].iter().cloned().cycle().take(100000).map(|v| std::char::from_digit(v, 10).unwrap()).collect_vec();
    // let n = x.len();

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

    let p = [ones + 1, ones.saturating_sub(1)];

    let mut rems = vec![vec![0i64; n]; 2];
    for i in 0..2 {
        if p[i] == 0 || p[i] == 1 {
            continue;
        }
        rems[i][n-1] = 1;
        for pos in (1..n).rev() {
            rems[i][pos-1] = 2i64 * rems[i][pos] % p[i];
        }
    }

    let mut sums = vec![0i64; 2];
    for i in 0..2 {
        for pos in 0..n {
            if x[pos] == '1' {
                sums[i] += rems[i][pos];
            }
        }
    }

    for i in 0..n {
        let (sel, sign) = if x[i] == '0' {
            (0, 1i64)
        } else {
            (1, -1i64)
        };
        if p[sel] == 0 {
            println!("0");
            continue;
        }
        let mut rem = (sums[sel] + sign * rems[sel][i]) % p[sel];
        let mut ans = 1;
        let mut div = rem.count_ones() as i64;
        while rem != 0 {
            rem %= div;
            div = rem.count_ones() as i64;
            ans += 1i64;
        }

        println!("{}", ans);
    }

}