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<vector>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<complex>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cassert>
using namespace std;

#define rep(i,n) for(int i=0;i<(int)n;i++)
#define each(i,c) for(__typeof(c.begin()) i=c.begin();i!=c.end();i++)
#define pb push_back
#define mp make_pair
#define all(c) c.begin(),c.end()
#define dbg(x) cerr<<__LINE__<<": "<<#x<<" = "<<(x)<<endl

typedef vector<int> vi;
typedef pair<int,int> pi;
typedef long long ll;
const int inf=(int)1e9;
const double EPS=1e-9, INF=1e12;

int n, w, ans;
int cnt[100010];

int main(){
	cin >> n >> w;
	for(int i = 1; i <= n; i++){
		for(int j = 1; j * j <= i; j++) if(i % j == 0){
			if(++cnt[j] == 2) ans++;
			if(j * j < i && ++cnt[i / j] == 2) ans++;
		}
		if(++cnt[i] == 2) ans++;
		
		if(i >= w) cout << ans << (i == n ? "\n" : " ");
		if(i < w) continue;
		
		int a = i - w + 1;
		for(int j = 1; j * j <= a; j++) if(a % j == 0){
			if(cnt[j]-- == 2) ans--;
			if(j * j < a && cnt[a / j]-- == 2) ans--;
		}
		if(cnt[a]-- == 2) ans--;
	}
	return 0;
}