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::{*, marker::*};

pub fn pow(n: usize, m: usize) -> usize {
    let mut ret = 1;
    let mut s = 2;
    let mut n = n;
    while n > 0 {
        if n & 1 > 0 {
            ret *= s;
            ret %= m;
        }
        s *= s;
        s %= m;
        n >>= 1;
    }
    ret
}

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

    let mut dp = vec![1_000_000_000usize; n*3];

    dp[0] = 0;
    dp[1] = 1;

    for i in 2..3*n {
        let t = i % (i as usize).count_ones() as usize;
        dp[i] = dp[t] + 1;
    }

    let mut ppcnt = 0;
    let mut x: Vec<char> = x;
    x.reverse();
    for i in 0..n {
        if x[i] == '1' {
            ppcnt += 1;
        }
    }

    let mut sums = vec![0; 3];
    for i in 0..n {
        if x[i] == '0' {
            continue;
        }
        for j in 0..3 {
            let tt = ppcnt as i32 - 1 + j as i32;
            if tt <= 0 {
                continue;
            }
            let t = pow(i, tt as usize);
            sums[j] += t;
            sums[j] %= tt as usize;
        }
    }

    for i in (0..n).rev() {
        if x[i] == '1' {
            let tt = ppcnt as i32 - 1;
            if tt == 0 {
                println!("0");
                continue;
            }
            let t = (sums[0] + tt as usize - pow(i, tt as usize)) % tt as usize;
            println!("{}", dp[t] + 1);
        } else {
            let tt = ppcnt as i32 + 1;
            let t = (sums[2] + pow(i, tt as usize)) % tt as usize;
            println!("{}", dp[t] + 1);
        }
    }
}