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)

struct REC(Vec<usize>);
impl REC {
    fn f(&mut self, x: usize) -> usize {
        if x == 0 {
            1
        } else if self.0[x] != !0 {
            self.0[x]
        } else {
            self.0[x] = self.f(x % x.count_ones() as usize) + 1;
            self.0[x]
        }
    }
}

fn main() {
    let n: usize = {
        let mut buf = String::new();
        std::io::stdin().read_line(&mut buf).unwrap();
        buf.trim_end().parse().unwrap()
    };
    let x: Vec<char> = {
        let mut buf = String::new();
        std::io::stdin().read_line(&mut buf).unwrap();
        buf.trim_end().chars().collect()
    };

    let mut m = 0;
    for i in 0..n {
        if x[i] == '1' {
            m += 1;
        }
    }
    let mut l = vec![0; n];
    let mut suml = 0;
    let mut r = vec![0; n];
    let mut sumr = 0;
    let mut p = 1;
    let mut q = 1;
    for i in (0..n).rev() {
        l[i] = p;
        r[i] = q;
        if x[i] == '1' {
            suml = (suml + p) % (m + 1);
            if m > 1 {
                sumr = (sumr + q) % (m - 1);
            }
        }
        p = p * 2 % (m + 1);
        if m > 1 {
            q = q * 2 % (m - 1);
        }
    }

    let mut rec = REC(vec![!0; n + 1]);

    for i in 0..n {
        if x[i] == '1' {
            if m == 1 {
                println!("0");
                continue;
            }
            println!("{}", rec.f((sumr + (m - 1) - r[i]) % (m - 1)));
        } else {
            println!("{}", rec.f((suml + l[i]) % (m + 1)));
        }
    }
}