Problem: Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.

A sequence of l integers b1, b2, ..., bl (1 ≤ b1 ≤ b2 ≤ ... ≤ bl ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally <image> for all i (1 ≤ i ≤ l - 1).

Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007 (109 + 7).

Input

The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).

Output

Output a single integer — the number of good sequences of length k modulo 1000000007 (109 + 7).

Examples

Input

3 2


Output

5


Input

6 4


Output

39


Input

2 1


Output

2

Note

In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3].

Solution (in JAVA):

Solution:
import java.lang.reflect.Array;
import java.util.*;
public class round_141b{
    public static void main(String args[]){
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int k=sc.nextInt();
        ArrayList<ArrayList<Integer>> al=new ArrayList<ArrayList<Integer>>();
        for(int i=0;i<n;i++){
            al.add(new ArrayList<Integer>());
            for(int j=1;j*j<=i+1;j++){
                if(i%j==0){
                    al.get(al.size()-1).add(j);
                    if(j*j!=i+1)al.get(al.size()-1).add((i+1)/j);
                }
            }
        }
        int a[][]=new int[k][n];
        Arrays.fill(a[0],1);
        int z=0;
        int mod=1000000007;
        for(int i=0;i<k-1;i++){
            for(int j=0;j<n;j++){
                z=j+1;
                while(z<=n){
                    a[i+1][z-1]=(a[i+1][z-1]+a[i][j])%mod;
                    z+=j+1;
                } 
            }
        }
        long ans=0;
        for(int i=0;i<n;i++){
            ans=(ans+a[k-1][i])%mod;
        }
        System.out.println(ans);
    }
}