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: Accepted)

#include <bits/stdc++.h>
#include <math.h>
#define lli long long int
#include <limits.h>
#include <fstream>

using namespace std;

lli gcd(lli a,lli b)
{
	if(a==0)return b;
	return gcd(b%a,a);
}

string s;

lli binpow(lli a,lli b,lli mod)
{
    if(b==0)
        return 1;
    lli ans=(binpow(a,b/2,mod))%mod;
    ans=(ans*ans)%mod;
    if(b%2)
        return (ans*a)%mod;
    else
        return ans;
}

lli solve(lli a)
{
    if(a==0)
        return 0;
    lli temp=a,ct=0;
    while(temp>0)
    {
        if(temp%2)
            ct++;
        temp/=2;
    }
    return 1+solve(a%ct);
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    lli i,j,tt=1;
    //cin>>tt;
    while(tt--)
    {
        lli n,o=0,z=0;
        cin>>n;
        cin>>s;
        for(i=0;i<n;i++)
        {
            if(s[i]=='1')
                o++;
            else
                z++;
        }
        vector<lli> possible;
        if(o==0)
        {
            possible.push_back(1);
            for(i=0;i<n;i++)
            {
                cout<<"1\n";
            }
            continue;
        }
        else if(z==0)
        {
            possible.push_back(n-1);
        }
        else
        {
            possible.push_back(o-1);
            possible.push_back(o+1);
        }
        lli rem[(int)possible.size()]={0};
        for(i=0;i<(int)possible.size();i++)
        {
            for(j=0;j<n;j++)
            {
                if(s[j]=='1' && possible[i]>0)
                {
                    rem[i]=(rem[i]+binpow(2LL,n-j-1,possible[i]))%possible[i];
                }
            }
        }
        for(i=0;i<n;i++)
        {
            if(s[i]=='1')
            {
                if(possible[0]==0)
                {
                    cout<<"0\n";
                    continue;
                }
                lli temp=binpow(2LL,n-1-i,possible[0]);
                temp=(rem[0]+possible[0]-temp)%possible[0];
                lli ans=1+solve(temp);
                cout<<ans<<"\n";
            }
            else
            {
                if(possible[possible.size()-1]==0)
                {
                    cout<<"0\n";
                    continue;
                }
                lli temp=binpow(2LL,n-1-i,possible[possible.size()-1]);
                temp=(rem[possible.size()-1]+temp)%possible[possible.size()-1];
                lli ans=1+solve(temp);
                cout<<ans<<"\n";
            }
        }
    }
    return 0;
}