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> 
using namespace std; 
#define ll long long int
#define pb push_back
ll cntone(ll n)
{
    ll cnt = 0,i;
    for(i=60;i>=0;i--)
        if((n>>i)&1)
            cnt++;
    return cnt;
}
ll power(ll x,ll y, ll p)  
{  
    ll res = 1;  
    x = x % p;    
    if (x == 0) return 0; 
    while (y > 0)  
    {  
        if (y & 1)  
            res = (res*x) % p;  
        y = y>>1; 
        x = (x*x) % p;  
    }  
    return res;  
}  
int main() 
{ 
    ll n;
    cin>>n;
    string x;
    cin>>x;
    ll i,j,ans = 0,cnt = 0,s = 0 , s2 = 0;
    for(i=n-1;i>=0;i--)
        if(x[i]=='1')
            cnt++;
    if(cnt==0)
    {
        for(i=0;i<n;i++)
            cout<<1<<"\n";
        return 0;
    }
    if(cnt==1)
    {
        for(i=0;i<n;i++)
        {
            if(x[i]=='1')
                cout<<0<<"\n";
            else
            {
                if(x[n-1]=='1')
                    cout<<2<<"\n";
                else
                    cout<<1<<"\n";
            }
        }
        return 0;
    }
    for(j=0,i=n-1;i>=0;i--,j++)
    {
        if(x[i]=='1'){
            s = (s+power(2,j,cnt+1))%(cnt+1);
            if(cnt-1!=0)
            s2 = (s2+power(2,j,cnt-1))%(cnt-1);
        }
        // cout<<s<<" "<<s2<<"\n";
    }
    for(j=0,i=n-1;i>=0;i--,j++)
    {
        ll w;
        ans = 1;
        if(x[j]=='1'){
            if(cnt-1!=0)
                w = (s2-power(2,i,cnt-1)+cnt-1)%(cnt-1);
            else{
                w = 0;
                ans = 0;
            }
        }
        else
            w = (s+power(2,i,cnt+1))%(cnt+1);
        // cout<<w<<" ";
        while(w!=0)
        {
            w = w%cntone(w);
            ans++;
        }
        cout<<ans<<"\n";
    }
    return 0; 
}