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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
#define FOR(i,l,r) for(i=l;i<r;i++)
#define REP(i,n) FOR(i,0,n)
#define ALL(x) x.begin(),x.end()
#define P pair<ll,ll>
#define F first
#define S second
ll A=0,B=0,N,x;
ll Q(ll&i,string&S,vector<ll>&F,vector<ll>&G){
  if(S.at(i)=='0')return (A+F.at(i))%(x+1);
  return ((B-G.at(i))%(x-1)+x-1)%(x-1);
}
ll f(ll m){
  if(m==0)return 0;
  ll i=0,n=m;
  while(n>0){
    if(n%2==1)i++;
    n/=2;
  }
  return 1+f(m%i);
}
signed main(){
  ll i,s=1;string S,T;cin>>N>>S;
  T=S;sort(ALL(T));
  x=upper_bound(ALL(T),'0')-T.begin();
  x=N-x;vector<ll>F(N);
  REP(i,N){
    F.at(N-1-i)=s;
    if(S.at(N-1-i)=='1')A+=s;
    s*=2;s%=x+1;A%=x+1;
  }
  s=1;vector<ll>G(N);
  if(x<=1){
    REP(i,N)G.at(i)=-1;
  }
 else REP(i,N){
    G.at(N-1-i)=s;
    if(S.at(N-1-i)=='1')B+=s;
    s*=2;s%=x-1;B%=x-1;
  }
  REP(i,N){
    T=S;if(T.at(i)=='0')T.at(i)='1';
    else T.at(i)='0';sort(ALL(T));
    if(T.at(N-1)=='0')cout<<0<<endl;
    else{
      ll RES=Q(i,S,F,G);
      cout<<1+f(RES)<<endl;
    }
  }
  return 0;
}