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 <cmath>
#include <string>


using namespace std;
int N;
int ModBinPow(int p, long long int a,int M, int Nnew){
  int res = 1;
  long long int fac = 1;
  for(int i = 0; i < Nnew; i++){
    if(a >> i & 1LL){
      res = (res * (int)pow(2, (fac << i))%M)%M;
    }
  }
  res %= M;
  return res;
}

string ChangeToBinString(long long int X, int Nnew){
  string res = new char[Nnew];
  for(int i = 0; i < Nnew; i++){
    if((X >> (Nnew - i - 1) & 1LL)){
      res.push_back('1');
    }else{
      res.push_back('0');
    }
  }
  return res;
}
int CountLetters(long long int X){
  long long int r = X;
  int cntr = 0;
  do{
    r /= 2;
    cntr++;
  }while(r > 0);
  return cntr;
}

long long int Popcount(long long int X, int Nnew){
  long long int mask = 1;
  long long int cntr = 0;
  for(int i = 0; i < Nnew; i++){
    if((X>>i & mask) == 1)
      cntr++;
  }
  return cntr;
}

int main(){
  cin >> N;
  string s;
  cin >> s;
  int p = 0;
  
  for(int i = 0; i < N; i++){
    char c = s[i];
    int a = c - '0';
    if(a)
      p++;
  }
  int cntr = 0;
  for(int i = 0; i < N; i++){
    string news;
    int newp;
    if(s[i] == '0')
      newp = p+1;
    else
      newp = p-1;
    for(int j = 0; j < N; j++){
      if(i == j){
        news[j] = s[j] == '0' ? '1' : '0';
      }else{
        news[j] = s[j];
      }
    }
    long long int temp = 0;
    int M = newp;
    int Nnew = N;
    cntr=0;
    do{
      for(int j = 0; j < Nnew; j++){
        if(news[Nnew-j-1] == '1'){
          temp = (temp + ModBinPow(2, j, M, CountLetters(j)))%M;
        }
      }
      cntr++;
      temp %= M;
      Nnew = CountLetters(temp);
      M = Popcount(temp, Nnew);
      news = ChangeToBinString(temp, Nnew);
    }while(temp != 0);
    cout << cntr<< endl;
  }
}