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 <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <iomanip>
#include <queue>
#include <deque>
#include <map>
#include <unordered_map>
#define rep(i,n) for(int i=0;i<n;i++)
#define repn(i,n) for(int i=1;i<=n;i++)
#define repr(e,x) for(auto& e:x)
using namespace std;
typedef long long ll;
typedef long double ld;
//typedef pair<int,int> P;
int const INF=1001001001;
ll const LINF=1001001001001001001;
ll const MOD=1000000007;

int N;
string X;

int popcount(int n){
    int res=0;
    while(n>0){
        res+=n%2;
        n/=2;
    }
    return res;
}

int f(int n,int cnt){
    if(n==0) return 1;
    n%=cnt;
    return f(n,popcount(n))+1;
}

int main(){
    cin>>N>>X;

    int cnt=0;
    rep(i,N) cnt+=X[i]-'0';

    int plus=0, minus=0;
    rep(i,N) plus=(plus*2+(X[i]-'0'))%(cnt+1);
    rep(i,N) minus=(minus*2+(X[i]-'0'))%(cnt-1);
    //cout<<plus<<' '<<minus<<'\n'<<endl;
    
    vector<int> tmpp(N+1); tmpp[N]=0;
    vector<int> tmpm(N+1); tmpm[N]=0;
    rep(i,N) tmpp[N-i-1]=(tmpp[N-i]*2+1)%(cnt+1);
    rep(i,N) tmpm[N-i-1]=(tmpm[N-i]*2+1)%(cnt-1);
    //rep(i,N) cout<<tmpp[i]<<endl;cout<<endl;
    //rep(i,N) cout<<tmpm[i]<<endl;cout<<endl;

    rep(i,N){
        int n;
        int diff;
        if(X[i]=='0'){
            diff=(tmpp[i]-tmpp[i+1]+cnt+1)%(cnt+1);
            n=(plus+diff+cnt+1)%(cnt+1);
        }else{
            diff=(tmpm[i]-tmpm[i+1]+cnt-1)%(cnt-1);
            n=(minus-diff+cnt-1)%(cnt-1);
        }
        //cout<<diff<<' '<<n<<endl;
        //cout<<n<<' '<<popcount(n)<<endl;
        int ans=f(n,popcount(n));
        cout<<ans<<endl;//cout<<endl;
    }
}