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 <algorithm>
#include <bitset>
#include <cassert>
#include <cctype>
#include <climits>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <fstream>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <valarray>
#include <vector>

#define EPS 1e-9
#define INF 1070000000LL
#define MOD 1000000007LL
#define fir first
#define foreach(it,X) for(__typeof((X).begin()) it=(X).begin();it!=(X).end();it++)
#define ite iterator
#define mp make_pair
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define pb push_back
#define sec second
#define sz(x) ((int)x.size())

using namespace std;

struct timer{
	time_t start;
	timer(){start=clock();}
	~timer(){cerr<<1.*(clock()-start)/CLOCKS_PER_SEC<<" secs"<<endl;}
};

typedef istringstream iss;
typedef long long ll;
typedef pair<int,int> pi;
typedef stringstream sst;
typedef vector<int> vi;

int N,W,w;
vi d[100010];

int main(){
	cin.tie(0);
	ios_base::sync_with_stdio(0);
	
	cin>>N>>W;
	w=W/2;
	vi ans;
	rep2(i,1,min(w+2,N-W+2)){
		ans.pb(i+W-1);
	}
	rep2(i,w+1,W){
		for(int j=i;j<=N;j+=i){
			d[j].pb(i);
		}
	}
	int res=ans[sz(ans)-1];
	rep2(i,w+2,N-W+2){
		int in=i+W-1,out=i-1;
		int hi=min(i-1,W-1);
		rep(j,sz(d[in]))if(d[in][j]<=hi)res++;
		rep(j,sz(d[out]))if(d[out][j]<=hi)res--;
		if(out<W)res++;
		ans.pb(res);
	}
	rep(i,sz(ans))cout<<ans[i]<<(i==sz(ans)-1?"\n":" ");
}