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>
#define ll long long
#define sz(x) (int)(x).size()
using namespace std;
//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
//uniform_int_distribution<int>(1000,10000)(rng)

int main()
{
    ios_base::sync_with_stdio(0); cin.tie(0);
    ll ans[200000+4];
    ans[0]=0;
    for (int i=1;i<200000+4;++i)
        ans[i]=ans[i%(__builtin_popcount(i))]+1;
    ll b=0,n;
    string s;
    cin>>n;
    cin>>s;
    for (auto c : s)
        if (c=='1')
            ++b;
    ll bin1[n+4],bin2[n+4],sc1[n+1+4],sc2[n+1+4],c1=0,c2=0,x;
    bin1[0]=1%(b-1);
    bin2[0]=1%(b+1);
    for (int i=1;i<n;++i)
    {
        bin1[i]=bin1[i-1]*2;
        bin1[i]%=(b-1);
        bin2[i]=bin2[i-1]*2;
        bin2[i]%=(b+1);
        //cout<<bin1[i]<<" "<<bin2[i]<<" <-\n";
    }
    sc1[n]=0;sc2[n]=0;
    for (int i=n-1;i+1;--i)
    {
        sc1[i]=sc1[i+1]+(s[i]-'0')*bin1[n-1-i];
        sc1[i]%=(b-1);
        sc2[i]=sc2[i+1]+(s[i]-'0')*bin2[n-1-i];
        sc2[i]%=(b+1);
        //cout<<sc1[i]<<" "<<sc2[i]<<"\n";
    }
    for (int i=0;i<n;++i)
    {
        //cout<<"! "<<c1<<" "<<c2<<"\n";
        if (b==1&&s[i]=='1')
            cout<<"0\n";
        else if (s[i]=='1')
        {
            x=c1*2;
            x+=(1-(s[i]-'0'));
            x%=(b-1);
            x*=bin1[n-1-i];
            x+=sc1[i+1];
            //cout<<ans[x%(b-1)]+1<<" "<<x%(b-1)<<"!\n";
            cout<<ans[x%(b-1)]+1<<"\n";
        }
        else
        {
            x=c2*2;
            x+=(1-(s[i]-'0'));
            //cout<<x<<" here\n";
            x%=(b+1);
            //cout<<x<<" here\n";
            x*=bin2[n-1-i];
            //cout<<x<<" here\n";
            x+=sc2[i+1];
            //cout<<x<<" here\n";
            //cout<<ans[x%(b+1)]+1<<" "<<x%(b+1)<<"!!\n";
            cout<<ans[x%(b+1)]+1<<"\n";
        }
        c1*=2;
        c1+=((s[i]-'0'));
        c1%=(b-1);
        c2*=2;
        c2+=((s[i]-'0'));
        c2%=(b+1);
    }
    return 0;
}