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>
using namespace std;
using ll=int64_t;

map<ll,ll>kiroku;

ll f(ll num,ll keta){
  if(num==0){
    cout <<0<<endl;
    return 0;
  }
  if(kiroku.count(num))return kiroku.at(num);
  ll ans=0;
  ll tmp=num;
  while(tmp){
    tmp%=keta;
    ans++;
    if(tmp==0){
      kiroku[num]=ans;
      return ans;
    }
    if(kiroku.count(tmp)){
      kiroku[num]=kiroku.at(tmp)+ans;
      return kiroku.at(num);
    }
    ll tmp2=tmp;
    keta=0;
    while(tmp2>0){
      tmp2/=2;
      keta++;
    }
    ll keta3=0;
    for (ll j=0;j<keta;j++){
      ll tmp5=tmp;
      if(tmp5>>j&1){
        keta3++;
      }
    }
    keta=keta3;
  }
}

int main(){
  ll n;cin>>n;
  string s;cin>>s;
  reverse(s.begin(),s.end());
  ll num=0;
  ll keta=0;
  for(ll i=0;i<n;i++){
    if(s.at(i)=='1'){
      num+=pow(2,i);
      keta++;
    }
  }
  for(ll i=n-1;i>=0;i--){
    ll keta2;
    ll tmp=num;
    if(s.at(i)=='1'){
      tmp-=pow(2,i);
      keta2=keta-1;
    }
    else{
      tmp+=pow(2,i);
      keta2=keta+1;
    }
    ll ans=0;
    ans=f(tmp,keta2);
    cout <<ans<<endl;
  }
}