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)

#![allow(
    non_snake_case,
    unused_variables,
    unused_assignments,
    unused_mut,
    unused_imports,
    dead_code
)]
//use proconio::fastout;
use proconio::input;
use proconio::marker::*;
//use std::collections::HashSet;
// use std::cmp::*;
//use std::collections::VecDeque;

fn main() {
    input! {
       N:usize, X:Chars
    }
    let X: Vec<u64> = X.iter().map(|&c| (c as u8 - '0' as u8) as u64).collect();
    let mut ones = 0;
    for i in 0..N {
        ones += X[i];
    }
    let mut amari_m = vec![1u64; N];
    let mut amari_p = vec![1u64; N];
    let mut amari_m_total = 0;
    let mut amari_p_total = 0;
    if ones > 1 {
        amari_m[0] = 1 % (ones - 1);
        amari_m_total += X[N - 1] * amari_m[0];
        for i in 1..N {
            // べき乗のあまりは循環するので、2^i % (one+-1)== 0となるiを見つける。(高々N==10万。)
            amari_m[i] = amari_m[i - 1] * 2 % (ones - 1);
            amari_m_total += X[N - i - 1] * amari_m[i];
        }
    }
    amari_p[0] = 1 % (ones + 1);
    amari_p_total += X[N - 1] * amari_p[0];
    for i in 1..N {
        // べき乗のあまりは循環するので、2^i % (one+-1)== 0となるiを見つける。(高々N==10万。)
        amari_p[i] = amari_p[i - 1] * 2 % (ones + 1);
        amari_p_total += X[N - i - 1] * amari_p[i];
    }
    //println!("ones={}", ones);
    //println!("m {},{:?}", amari_m_total, amari_m);
    //println!("p {},{:?}", amari_p_total, amari_p);
    for i in 0..N {
        if ones == 1 && X[i] == 1 {
            println!("0");
            continue;
        }
        let mut t = if X[i] == 1 {
            //       println!("{} - {}, {}", amari_m_total, amari_m[N - i - 1], ones - 1);
            (amari_m_total - amari_m[N - i - 1]) % (ones - 1)
        } else {
            //     println!("{} + {}, {}", amari_p_total, amari_p[N - i - 1], ones + 1);
            (amari_p_total + amari_p[N - i - 1]) % (ones + 1)
        };
        let mut ans = 1;
        while t > 0 {
            ans += 1;

            let mut ones2 = 0;
            let mut t2 = t;
            while t2 > 0 {
                ones2 += t2 & 1;
                t2 = t2 >> 1;
            }
            t = t % ones2;
        }
        println!("{}", ans);
    }
}