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 std::cmp::*;
use std::collections::*;
use std::io::*;
use std::str::FromStr;
fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}
fn pcnt(x: i32) -> i32 {
    x.count_ones() as i32
}
fn f(x: i32) -> i32 {
    if x == 0 {
        return 0;
    }
    f(x % pcnt(x)) + 1
}
fn main() {
    let n: usize = read();
    let s: String = read();
    let mut s: Vec<char> = s.chars().collect();
    let mut x: Vec<i32> = vec![0; n];
    for i in 0..n {
        x[i] = s[i] as i32 - '0' as i32;
    }
    let mut pc = 0;
    for i in 0..n {
        pc += x[i];
    }
    let mut ans: Vec<i32> = vec![0; n];
    for b in 0..2 {
        let mut npc = pc;
        if b == 0 {
            npc += 1;
        } else {
            npc -= 1;
        }
        if npc <= 0 {
            continue;
        }
        let mut r0 = 0;
        for i in 0..n {
            r0 = (r0 * 2) % npc;
            r0 += x[i];
        }
        let mut k = 1;
        for i in 0..n {
            if x[n - 1 - i] == b {
                let mut r = r0;
                if (b == 0) {
                    r = (r + k) % npc;
                } else {
                    r = (r - k + npc) % npc;
                }
                ans[n - 1 - i] = f(r) + 1;
            }
            k = (k * 2) % npc
        }
    }
    for i in 0..n {
        println!("{}", ans[i]);
    }
}