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>
#include<sstream>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<algorithm>
#include<functional>
#include<utility>
#include<cmath>
#include<complex>
#include<iomanip>

using namespace std;

#define REP(i,s,e) for(int i=int(s);i<=int(e);i++)
#define rep(i,n) for(int i=0;i<int(n);i++)
#define pi 3.14159265358979

int a[100010];

int main(void){

	int n,w;
	cin >> n >> w;

	for(int i=1;i<=w/2;i++)
    a[i]=w-1+i;

	for(int i=w/2+1;i<=n-w+1;i++)
	a[i]=w+w/2;

	for(int j=w/2+1;j<=w-1;j++){
		for(int k=3*j;k<=n;k+=j){
			for(int i=k-w+1;i<=k-j;i++)
				a[i]+=1;
		}
	}

	for(int i=1;i<=n-w+1;i++){
    if(i!=n-w+1) cout << a[i] << ' ';
    else cout << a[i] << endl;

	}

	return 0;
}