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: Wrong Answer)

fn main() {
    proconio::input! {
        n: usize,
        x: proconio::marker::Chars,
    }
    fn power_of_2_mod_bit_count(n: u64, bit_count: u64) -> Vec<u64> {
        let mut v = vec![0; n as usize];
        let mut tmp = 1;
        for i in 0..n {
            tmp %= bit_count;
            v[n as usize - i as usize - 1] = tmp;
            tmp <<= 1;
        }
        v
    }
    fn f(x: u64) -> u64 {
        let count = x.count_ones() as u64;
        x % count
    }
    let count_ones = x.iter().filter(|x| **x == '1').count() as u64;
    println!("{}", count_ones);
    //let ms_plus1 = power_of_2_mod_bit_count(n as u64, count_ones + 1);
    //if count_ones == 0 {
    //    for _ in 0..n {
    //        println!("1");
    //    }
    //    return;
    //}
    //let ms_minus1 = power_of_2_mod_bit_count(n as u64, count_ones - 1);
    //let m_plus1 = ms_plus1
    //    .iter()
    //    .zip(x.iter())
    //    .map(|(i, c)| if *c == '1' { *i } else { 0 })
    //    .sum::<u64>()
    //    % (count_ones + 1);
    //let m_minus1 = ms_minus1
    //    .iter()
    //    .zip(x.iter())
    //    .map(|(i, c)| if *c == '1' { *i } else { 0 })
    //    .sum::<u64>()
    //    % (count_ones - 1);
    //for i in 0..n {
    //    let (count_ones, is_minus1) = if x[i] == '1' {
    //        (count_ones - 1, true)
    //    } else {
    //        (count_ones + 1, false)
    //    };
    //    let mut x = if is_minus1 {
    //        (m_minus1 + count_ones - ms_minus1[i]) % count_ones
    //    } else {
    //        (m_plus1 + ms_plus1[i]) % count_ones
    //    };
    //    let mut count = 1usize;
    //    while x != 0 {
    //        x = f(x);
    //        count += 1;
    //    }
    //    println!("{}", count);
    //}
}