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: Wrong Answer)

fn main() {
    proconio::input! {
        _n: usize,
        x: String,
    }

    let mut xx = 0;
    let mut xi = 1;
    for c in x.chars().rev() {
        let c = if c == '1' { 1 } else { 0 };
        xx += c * xi;
        xi *= 2;
    }
    for c in x.chars() {
        xi /= 2;
        let xx = if c == '1' { xx - xi } else { xx + xi };
        println!("{}", pop_pop_count(xx));
    }
}

fn pop_pop_count(mut v: u64) -> u64 {
    let mut r = 0;
    while v > 0 {
        v = v % pop_count(v);
        r += 1;
    }
    r
}

fn pop_count(v :u64) -> u64
{
   let mut count = (v & 0x5555555555555555) + ((v >>  1) & 0x5555555555555555);
   count = (count & 0x3333333333333333) + ((count >>  2) & 0x3333333333333333);
   count = (count & 0x0f0f0f0f0f0f0f0f) + ((count >>  4) & 0x0f0f0f0f0f0f0f0f);
   count = (count & 0x00ff00ff00ff00ff) + ((count >>  8) & 0x00ff00ff00ff00ff);
   count = (count & 0x0000ffff0000ffff) + ((count >> 16) & 0x0000ffff0000ffff);
/*count=*/ (count & 0x00000000ffffffff) + ((count >> 32) & 0x00000000ffffffff)
}