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)

fn main() {
    proconio::input! {
        n: usize,
        x: String,
    }
    let x=x.as_bytes();
    let b=x.iter().filter(|&&c|c!=b'0').count();
    let mut d=[0;2];
    let mut r=vec![vec![0;n];2];
    for i in 0..2 {
        if i==0 && b<2 {
            continue;
        }
        let b=b+i*2-1;
        let mut v=0;
        for &x in x {
            v=(v*2+(if x==b'0' {0} else {1}))%b;
        }
        d[i]=v;
        r[i][0]=1;
        for k in 1..n {
            r[i][k]=r[i][k-1]*2%b;
        }
        r[i].reverse();
    }
    for (i,&x) in x.iter().enumerate() {
        let v=if x==b'0' {0} else {1};
        let b=if v==0 {b+1} else {b-1};
        if b==0 {
            println!("0");
            continue;
        }
        let mut a=if v==0 {
            (d[1]+r[1][i])%b
        }
        else {
            (d[0]+b-r[0][i])%b
        };
        let mut c=1;
        while a>0 {
            a%=a.count_ones()as usize;
            c+=1;
        }
        println!("{}",c);
    }
}