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 proconio::{input, fastout};

fn ppc(mut x: u64) -> u64 {
    let mut c = 0;
    while x > 0 {
        if x & 1 != 0 {c += 1}
        x >>= 1;
    }
    c
}

fn check(mut x: u64) -> u64 {
    let mut c = 0;
    while x > 0 {
      x %= ppc(x);
      c += 1;
    }
    c
}

fn pow2m(mut x: u64, m: u64) -> u64 {
    let mut d: u64 = 2;
    let mut ans: u64 = 1;
    while x > 0 {
        if x & 1 != 0 {
            ans *= d;
            ans %= m;
        }
        d *= d;
        d %= m;
        x >>= 1;
    }
    ans
}

#[fastout]
fn main() {
    input! {
        n: usize,
        mut x: String,
    }
    let mut ppc = 0;
    for c in x.chars() {
        if c == '1' {ppc += 1}
    }
    let ppcp = ppc + 1;
    let ppcm;
    if ppc > 1 {
        ppcm = ppc - 1;
    } else {
        ppcm = ppc;
    }
    // xp,xm = x % ppcp,ppcm
    let mut xp = 0;
    let mut xm = 0;
    let mut dp = 1;
    let mut dm = 1;
    for c in x.chars().rev() {
        if c == '1' {
            xp += dp;
            if xp >= ppcp { xp -= ppcp }
            if ppc > 1 {
                xm += dm;
                if xm >= ppcm { xm -= ppcm }
            }
        }
        dp *= 2;
        dp %= ppcp;
        if ppc > 1 {
            dm *= 2;
            dm %= ppcm;
        }
    }
    let mut i = n as u64;
    for c in x.chars() {
        i -= 1;
        let mut xt;
        if c == '0' {
            xt = xp + pow2m(i, ppcp);
            if xt >= ppcp {xt -= ppcp}
            if xt == 0 {
                println!("1");
            } else {
                println!("{}", check(xt) + 1);
            }
        } else {
            if ppc == 1 {
                println!("0");
            } else {
                xt = xm + ppcm - pow2m(i, ppcm);
                xt %= ppcm;
                if xt == 0 {
                    println!("1");
                } else {
                    println!("{}", check(xt) + 1);
                }
            }
        }
    }
}