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 C++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
using namespace std;

int deal(int x){
    int res=0;
    while(x){
        res++;
        x=x%__builtin_popcount(x);
    }
    return res;
}

int n,p1=-1,p2=1,x1,x2;
char s[200005];
int pw1[200005],pw2[200005];

int main(){

    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    cin>>n>>s+1;
    for(int i=1;i<=n;i++){
        p1+=s[i]=='1';
        p2+=s[i]=='1';
    }
    pw1[0]=pw2[0]=1;
    for(int i=1;i<=n;i++){
        if(p1)x1=(x1<<1|(s[i]=='1'))%p1;
        x2=(x2<<1|(s[i]=='1'))%p2;
        if(p1)pw1[i]=(pw1[i-1]<<1)%p1;
        pw2[i]=(pw2[i-1]<<1)%p2;
    }
    for(int i=1;i<=n;i++){
        if(s[i]=='1'){
            cout<<deal((x1-pw1[n-i]+p1)%p1)+1<<endl;
        }else{
            cout<<deal((x2+pw2[n-i]+p2)%p2)+1<<endl;
        }
    }

    return 0;
}