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<stdio.h>
#include<algorithm>
using namespace std;
pair<int,int> event[5000000];
int now[100001];
int main(){
	int a,b;
	scanf("%d%d",&a,&b);
	int at=0;
	for(int i=1;i<=a;i++){
		event[at++]=make_pair(i-b+1,i);
		event[at++]=make_pair(i+1,-i);
		for(int j=1;j*j<=i;j++){
			if(i%j==0){
				int c=i/j;
				if(j<b){
					event[at++]=make_pair(i-b+1,j);
					event[at++]=make_pair(i-j+1,-j);
					event[at++]=make_pair(i+j-b+1,j);
					event[at++]=make_pair(i+1,-j);
				}
				if(c<b){
					event[at++]=make_pair(i-b+1,c);
					event[at++]=make_pair(i-c+1,-c);
					event[at++]=make_pair(i+c-b+1,c);
					event[at++]=make_pair(i+1,-c);
				}
			}
		}
	}
	std::sort(event,event+at);
	int ret=0;
	int right=0;
	int left=0;
	for(int i=1;i<=a-b+1;i++){
		while(event[right].first<=i){
			if(event[right].second>0){
				now[event[right].second]++;
				if(now[event[right].second]==1)ret++;
			}
			right++;
		}
		while(event[left].first<=i){
			if(event[left].second<0){
				now[-event[left].second]--;
				if(now[-event[left].second]==0)ret--;
			}
			left++;
		}
		//for(int j=1;j<=10;j++)printf("%d ",now[j]);

		printf("%d",ret);
		if(i<a-b+1)printf(" ");
		else printf("\n");
	}
}