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

int popcount(int n){
  int temp;
  int amari;
  int count=0;
  temp=n;

  while(temp!=0){
    amari=temp%2;
    temp=temp/2;
    if(amari==1)count++;
  }
  return(count);
}

int main(){
  long long N;
  string X;
  long long stX;
  long long xi;
  int count;
  cin >> N;
  cin >> X;
  stX=stoi(X,0,2);
  cout << "stX:" << stX << endl;

  for(int i=1;i<=N;i++){
    count=0;
    xi = stX ^ (1<<(N-i));//反転整数
    cout << "xi:" << xi << endl;
    while(xi!=0){
      xi=xi%popcount(xi);
      count++;
    }
    cout << count << endl;
  }

  return(0);
}