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.util.Arrays;
import java.util.Scanner;

public class C {
    
static  int n;
static int k;
static int mod=1000000007;
static long dp[][];
    public static void main(String[] args) {
                
            Scanner x=new Scanner (System.in);
             n=x.nextInt();
            k=x.nextInt();
                
     ///  1 2 3 4 5 6         2000 
            
            dp=new long[2001][2001];
            for(int i=0;i<2001;i++){
                for(int j=0;j<2001;j++){
                    dp[i][j]=-1;
                }
            }
            
            System.out.println(solve(0,1));
    }
    
    
    public static long solve(int len,int numnow){
        
        if(len==k){return 1;}
                
        if(dp[len][numnow]!=-1){return dp[len][numnow];}
                long b=0;   
                for(int i=numnow;i<=n;i+=numnow){
                    
                    if(i%numnow==0){
                        b=b+solve(len+1,i);                     
                    }
                }
                dp[len][numnow]=b%mod;
                return b%mod;
            }   

}