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)

#![allow(unused_imports)]
#![allow(clippy::many_single_char_names)]

// language: Rust(1.42.0)
// check available crates on AtCoder at "https://atcoder.jp/contests/language-test-202001"
// My Library Repositry is at "https://github.com/Loptall/prcn_lib"

/*
    青perfほしいよ〜！
*/

// from std...
use std::cmp::{
    max, min, Ordering,
    Ordering::{Equal, Greater, Less},
    Reverse,
};
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, HashMap, HashSet, VecDeque};
use std::convert::TryInto;
use std::mem::swap;
use std::num::{NonZeroU32, ParseIntError};
use std::ops::{
    Add, AddAssign, BitAnd, BitAndAssign, BitOr, BitOrAssign, BitXor, BitXorAssign, Div, DivAssign,
    Index, IndexMut, Mul, MulAssign, Neg, Rem, RemAssign, Shl, ShlAssign, Shr, ShrAssign, Sub,
    SubAssign,
};
use std::process::exit;
use std::{f32, f64, i128, i16, i32, i64, i8, isize, u128, u16, u32, u64, u8, usize};

// Dep' crates are...
use itertools::*;
use itertools_num::*;
use lazy_static::lazy_static;
use maplit::{btreemap, btreeset, hashmap, hashset};
use num_bigint::{BigInt, BigUint};
use num_complex::Complex;
use num_integer::{binomial, gcd, lcm, multinomial, Integer};
use num_rational::Rational;
use num_traits::{clamp, one, pow, zero, Num, NumAssignOps, NumOps, One, Pow, Zero};
use permutohedron::Heap;
use proconio::{
    fastout, input, is_stdin_empty,
    marker::{Bytes, Chars, Isize1, Usize1},
};
use rand::random;

pub const MOD10E9_7: usize = 1000000007;
pub const MOD99_: usize = 998244353;
pub const MAX: usize = std::usize::MAX;
pub const PI: f64 = f64::consts::PI;
pub const ASCII_A: u8 = 65;

// code...
// #[fastout]
fn main() {
    input! {
        n: usize,
        x: Chars
    }

    // はじめ X の pop_count
    let mut p = 0usize;
    let x = x
        .iter()
        .map(|x| {
            if *x == '1' {
                p += 1;
                1u8
            } else {
                0u8
            }
        })
        .collect_vec();
    let p = p;

    let mut ans = vec![0usize; n];

    for b in 0..2 {
        let mut np = p;
        if b == 0 {
            np += 1;
        } else {
            np -= 1;
        }
        let np = np;

        let mut r = 0usize;
        for i in 0..n {
            r *= 2;
            r %= np;
            r += x[i] as usize;
        }

        let mut k = 1;
        for i in (0..n).rev() {
            if x[i] == b {
                let mut t = r;
                if b == 0 {
                    t += k;
                    t %= np;
                } else {
                    t += np - k;
                    t %= np;
                }
                ans[i] = f(t) + 1;
            }
            k *= 2;
            k %= np;
        }
    }

    for i in ans {
        println!("{}", i);
    }
}

fn f(n: usize) -> usize {
    if n == 0 {
        0
    } else {
        f(n % n.count_ones() as usize) + 1
    }
}