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)

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

    let bp = x.match_indices('1').count() as u32;
    let p = if bp == 1 { [2, 1] } else { [bp + 1, bp - 1] };

    let xr: Vec<char> = x.chars().rev().collect();

    let mut pows = vec![[1, 1]; n];
    let mut xv = if xr[0] == '1' { [1, 1] } else { [0, 0] };
    for i in 1..n {
        for j in 0..=1 {
            pows[i][j] = (pows[i - 1][j] * 2) % p[j];
            if xr[i] == '1' {
                xv[j] += pows[i][j];
                xv[j] %= p[j];
            }
        }
    }

    for i in (0..n).rev() {
        if bp == 1 && xr[i] == '1' {
            println!("0");
        } else {
            let xi = if xr[i] == '0' {
                (xv[0] + pows[i][0]) % p[0]
            } else {
                let p = if xv[1] >= pows[i][1] { 0 } else { p[1] };
                xv[1] + p - pows[i][1]
            };
            println!("{}", 1 + pop_pop_count(xi));
        }
    }
}

fn pop_pop_count(mut v: u32) -> u32 {
    let mut r = 0;
    while v > 0 {
        v %= v.count_ones() as u32;
        r += 1;
    }
    r
}