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)

// -*- coding:utf-8-unix -*-

use proconio::input;

fn pcnt(x: i32) -> i32 {
    x.count_ones() as i32
}

fn f(x: i32) -> i32 {
    if x == 0 {
        0
    } else {
        f(x % pcnt(x)) + 1
    }
}

fn main() {
    input! {
        n: u64,
        s: String,
    }

    let x: Vec<u8> = s.chars().map(|c| if c == '0' { 0 } else { 1 }).collect();
    let pc: i32 = x.iter().filter(|&i| *i == 1).count() as i32;
    let mut ans = Vec::with_capacity(n as usize);
    for _i in 0..n {
        ans.push(0);
    }

    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 as usize] as i32;
        }
        let mut k = 1;
        for i in (0..n).rev() {
            // `ii` 桁目を見ている際には `2^ii` が求まっている
            if x[i as usize] == b as u8 {
                let mut r = r0;
                if b == 0 {
                    r = (r + k) % npc;
                } else {
                    r = (npc + r - k) % npc;
                }
                ans[i as usize] = f(r) + 1;
            }
            k = (k * 2) % npc;
        }
    }

    for i in 0..n {
        println!("{}", ans[i as usize]);
    }
}