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 <cstdio>
#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#include <numeric>
#include <climits>
#include <cfloat>
#include <functional>
using namespace std;

int main()
{
    int n, w;
    cin >> n >> w;

    vector<int> a(n+1, 0);
    for(int i=1; i<=w/2; ++i)
        ++ a[i+1];
    for(int i=w/2+1; i<w; ++i){
        for(int j=3*i; j<=n; j+=i){
            ++ a[j-w+1];
            -- a[j-i+1];
        }
    }

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

    return 0;
}