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::{input, fastout};

fn make_dp(n:usize) -> Vec<usize> {
    let mut dp = vec![0; n];
    for i in 1 .. n {
        let bits = i.count_ones() as usize;
        dp[i] = dp[i % bits] + 1;
    }
    dp
}

fn source_dp(n:usize, m:usize, x:&[u8]) -> (usize, usize, Vec<usize>) {
    let mut dp = vec![1;n];
    let mut score = 0;

    if m == 0 {
        return (m, score, dp);
    }

    for i in (0 ..= n - 2).rev() {
        dp[i] = (dp[i + 1] * 2) % m;
    }
    
    for i in 0 .. n {
        if x[i] == b'1' {
            score = (score + dp[i]) % m;
        } 
    }
    (m, score + m, dp)
}

#[fastout]
fn main() {
    input!{
        n:usize,
        x:String,
    }

    let dp = make_dp(n);
    let x:Vec<u8> = x.as_bytes().to_vec();
    let popcount = x.iter().filter(|&b| *b == b'1').count();
    let (mod1, score1, dp1) = source_dp(n, popcount + 1, &x);
    let (mod2, score2, dp2) = source_dp(n, popcount - 1, &x);
    for i in 0 .. n {
        if x[i] == b'0' {
            println!("{}", dp[(score1 + dp1[i]) % mod1] + 1);
        } else if popcount == 1 {
            println!("{}", 0);
        } else {
            println!("{}", dp[(score2 - dp2[i]) % mod2] + 1);
        }
    }
}