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::input;

fn main() {
    input! {
        n: usize,
        x: String
    };
    let s = x.as_bytes();
    let mut cnt = 0;
    for &c in s {
        if c == b'1' { cnt += 1 }
    }
    let mut dp = vec![0; n+1];
    for i in 1..=n as i32 {
        let d = i.count_ones() as i32;
        dp[i as usize] = dp[(i%d) as usize] + 1;
    }
    let mut res = vec![0; n];
    for &d in [-1, 1].iter() {
        let m = cnt + d;
        if m <= 0 || (n as i32) < m { continue }
        let mut p2 = vec![1; n];
        let mut sum = 0;
        for i in (0..n-1).rev() {
            p2[i] = 2*p2[i+1] % m;
        }
        for i in 0..n {
            if s[i] == b'1' { sum = (sum + p2[i]) % m; }
        }
        for i in 0..n {
            if d == -1 {
                if s[i] == b'0' { continue }
                res[i] = dp[((sum+m-p2[i])%m) as usize] + 1;
            } else {
                if s[i] == b'1' { continue }
                res[i] = dp[((sum+p2[i])%m) as usize] + 1;
            }
        }
    }
    for &d in &res {
        println!("{}", d);
    }
}