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: Runtime Error)

use proconio::{fastout, input};

#[fastout]
fn main() {
    let x: String;
    input! {n: usize, x: String};
    let mut bin = 0b0;
    let mut k = 0b1;

    for i in x.chars().rev() {
        if i == '1' {
            bin = bin | k;
        }
        // println!("{}", bin);

        k = k << 1;
    }

    let mut v = Vec::new();

    for i in 0..n {
        let mut bin2 = bin ^ (1 << i);
        // 二進数 % 1の数
        let mut count = 0;
        let mut l = bin2;
        while l != 0 {
            let mut sum = 0;
            let mut m = l;
            while m != 0 {
                if m % 2 == 1 {
                    sum += 1
                }
                m /= 2;
            }
            l = l % sum;
            count += 1;
        }
        v.push(count);
    }

    for i in v.iter().rev() {
        println!("{}", i)
    }
}