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<stdio.h>
int main(){
  int n,count=0,i;
  char x[200002];
  scanf("%d",&n);
  scanf("%*c%s",x);
  for(i=0;i<n;i++){
    if(x[i]=='1')
      count++;
  }
  int popcount[count+2];
  popcount[0]=0;
  for(i=1;i<200001;i++){
    int temp=i,check=0;
    while(temp!=0){
      if(temp%2==1)
        check++;
      temp/=2;
    }
    popcount[i]=popcount[i%check]+1;
  }
  int mod[200002][2];
  int num[2]={0,0};
  for(i=n-1;i>=0;i--){
    if(i==n-1){
      mod[i][0]=1;
      mod[i][1]=1;
    }
    else{
      mod[i][0]=2*mod[i+1][0];
      mod[i][0]%=count+1;
      mod[i][1]=2*mod[i+1][1];
      mod[i][1]%=count-1;
    }
    if(x[i]=='1'){
      num[0]+=mod[i][0];
      num[0]%=count+1;
      num[1]+=mod[i][1];
      num[1]%=count-1;
    }
  }
  for(i=0;i<n;i++){
    int ans;
    if(x[i]=='1' && count-1==0)
      ans=0;
    else if(x[i]=='1')
      ans=popcount[(num[1]-mod[i][1]+count-1)%(count-1)]+1;
    else
      ans=popcount[(num[0]+mod[i][0])%(count+1)]+1;
    printf("%d\n",ans);
  }
  return 0;
}