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++ (Clang 10.0.0) (judge verdict: Accepted)

#include <iostream>
#include <string>
using namespace std;
int main(){
  int N;
  cin>>N;
  string s;
  cin>>s;
  int num=0;
  for(int i=0;i<N;i++) if(s[i]=='1') num++;
  int amari[200000][2];
  amari[0][0]=1; amari[0][1]=1;
  int judge[2]={0,0};
  for(int i=1;i<N;i++){
    if(num!=1) amari[i][0]=(amari[i-1][0]*2)%(num-1);
    amari[i][1]=(amari[i-1][1]*2)%(num+1);
  }
  for(int i=0;i<N;i++){
    if(s[N-i-1]=='1'){
      if(num!=1) judge[0]=(judge[0]+amari[i][0])%(num-1);
      judge[1]=(judge[1]+amari[i][1])%(num+1);
    }
  }
  for(int i=0;i<N;i++){
    int k;
    if(s[i]=='1'){
      if(num!=1) k=(judge[0]+num-1-amari[N-i-1][0])%(num-1);
      else k=-1;
    }
    else k=(judge[1]+amari[N-i-1][1])%(num+1);
    if(k!=-1){
    int answer=1;
    while(k>0){
      int l=0;
      long long zwei=1;
      while(zwei<=k){
        if((k%(zwei*2))/zwei==1) l++;
        zwei*=2;
      }
      k%=l;
      answer++;
    }
    cout<<answer<<endl;
    }
    if(k==-1) cout<<0<<endl;
  }
  
  return 0;
}