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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

int popcnt(long long n){
  int c=0;
  while(n>0){
    c+=(n%2);
    n/=2;
  }
  return c;
}

int main(){
  int N; cin >> N;
  string X; cin >> X;
  int px=0;
  for(int i=0; i<N; i++){
    if(X[i]=='1') px++;
  }
  
  vector<long long> resplus(N,0),resminus(N,0);
  resplus[N-1]=1;
  for(int i=N-1; i>0; i--){
    resplus[i-1]=(2*resplus[i])%(px+1);
  }
  if(px>1){
    resminus[N-1]=1;
    for(int i=N-1; i>0; i--){
      resminus[i-1]=(2*resminus[i])%(px-1);
    }
  }
  
  for(int i=0; i<N; i++){
    int count=1;
    if(X[i]=='1')
    {
      //1->0なのでpx--でスタート
      if(px>1){
        int temp=0;
        for(int j=0; j<N; j++){
          if(i!=j && X[j]=='1'){
            temp+=resminus[j];
            temp%=(px-1);
          }
        }
        while(temp>0){
          temp%=popcnt(temp);
          count++;
        }
      }
    }
    else
    {
      //0->1なのでpx++でスタート
      int temp=0;
      for(int j=0; j<N; j++){
        if(i==j || X[j]=='1'){
          temp+=resplus[j];
          temp%=(px+1);
        }
      }
      while(temp>0){
        temp%=popcnt(temp);
        count++;
      }
    }
    cout << count << endl;
  }
}