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: Wrong Answer)

    #include <bits/stdc++.h> 
/*#ifndef M_PI
    #define M_PI 3.14159265358979323846
#endif
*/

using namespace std;
#define m 1000000001
#define rep(i,a,b) for(int i=a;i<b;++i)
#define repb(i,a,b) for(int i=a;i>=b;--i)
#define repv(arr)  for(auto it=arr[i].begin();it!=arr[i].end();++it)  
#define vi vector<int>
#define vb vector<bool>
#define vs vector<string>
#define vl vector<long long int>
#define ssortA(arr)   stable_sort(arr.begin(),arr.end())
#define search(arr,c)  binary_search(arr.begin(),arr.end(),c)
#define pb   push_back
#define ll long long
#define mp make_pair 
#define LCM(a,b) boost::math::lcm(a,b)
#define cl_bf cin.ignore(numeric_limits<streamsize>::max(), '\n');
int main(){
    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int n;
    cin>>n;
    vector<char> arr(n);
    vi i1,i2;
    int c=0;
    rep(i,0,n){
        cin>>arr[i];
        if(arr[i]=='1')
        ++c;
    }
    //cout<<c<<endl;
    int x=c-1;
    if(x==0){
        rep(i,0,n){
            if(arr[i]=='0')cout<<1<<endl;
            else{
                cout<<0<<endl;
            }
        }
        return 0;
    }
    int y=c+1;
    vi b(200001,0);
    rep(i,1,200001){
        int j=i;
        int c1=0;
        while(j!=0){
            if(j%2!=0)
            ++c1;
            j/=2;
        }
        b[i]=1+b[i%c1];
        
        /*if(i>199900){
        cout<<b[i]<<" ";
        }*/
        
    }
    //cout<<b[15]<<endl;
    vi p1(n),p2(n);
    p1[0] = 1 % x;  
    for (int i = 1; i < n; i++)  
    {  
        p1[i] = p1[i - 1] * (2 % x);  
        p1[i] %= x;  
       // cout<<p1[i]<<" ";
    }  
    //cout<<endl;
    p2[0] = 1 % y;  
    for (int i = 1; i < n; i++)  
    {  
        p2[i] = p2[i - 1] * (2 % y);  
        p2[i] %= y;  
    } 
    ll sum1=0,sum2=0;
    rep(i,0,n){
        if(arr[i]=='1'){
            sum1+=p1[n-1-i];
            sum1=sum1%x;
        }
    }
    rep(i,0,n){
        if(arr[i]=='1'){
            sum2+=p2[n-1-i];
            sum2=sum2%y;
        }
    }
    rep(i,0,n){
        if(arr[i]=='1'){
            int g=sum1;
            int e=p1[n-1-i]%x;
            //cout<<e<<" "<<sum1<<endl;
            g=(sum1-e+x)%x;
            int f=1+b[g];
            cout<<f<<endl;
        }
        else{
            int g=sum2;
            int e=p2[n-1-i]%y;
            g=(sum2+e+y)%y;
            int f=1+b[g];
            cout<<f<<endl;
        }
    }
    //cout<<"hi"<<endl;
}