Sliding GCD

Problem Statement

For the set S of natural numbers, let f (S) be the number of elements in the set \\ {GCD (T) | T ⊆ S, T is not empty \\}.

Here, GCD (T) is the greatest integer that divides all the numbers contained in T.
In particular, note that GCD (\\ {a \\}) = a when T consists of only one integer a.

Find f (\\ {i, i + 1, ..., i + W --1 \\}) for i = 1, 2, ..., N --W + 1.

Constraints

* 1 ≤ W ≤ N ≤ 10 ^ 5

Input

Input follows the following format. All given numbers are integers.


N W

Output

The value of f (\\ {i, i + 1, ..., i + W-1 \\}) when i = 1, 2, ..., N-W + 1 is separated by a space. Print on a line.

Examples

Input

10 2


Output

2 3 3 3 3 3 3 3 3


Input

30 7


Output

7 8 9 10 10 11 11 11 11 12 11 12 10 12 12 11 10 12 12 12 10 11 11 13
#include<iostream>

using namespace std;

int main(){
  int N,W;
  cin>>N>>W;
  int c=0;
  int a[100001]={};
  for(int i=1;i<=N;i++){
    if(W<2*i&&i<W){
      a[i+1]+=i*3<=i+1+W-1;
      for(int j=i;j<=2*N;j+=i){
	int x=j-W+1;
	if(x>i+1&&x<=N){
	  a[x]++;
	}
	if(i+1<j+1&&j+1<=N){
	  a[j+1]--;
	}
      }
    }else if(i<W){
      a[i+1]++;
    }
  }
  for(int i=1;;i++){
    cout<<c+W;
    if(i==N-W+1){
      cout<<endl;
      break;
    }else{
      cout<<' ';
      c+=a[i+1];
    }
  }
}

      