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.io.*;
import java.util.*;
public class Solution
{
    static int MOD=(int)1e9+7;
    static long dp[][];
    static int N;
    public static void main(String ag[])
    {
        Scanner sc=new Scanner(System.in);
        int i,j,k;
        N=sc.nextInt();
        int K=sc.nextInt();
        dp=new long[N+1][K+1];
        for(i=0;i<=N;i++)
        Arrays.fill(dp[i],-1);
        
        System.out.println(find(1,K));
        
    }
    public static long find(int prev,int count)
    {
        if(count==0)
        return 1;
        
        if(dp[prev][count]!=-1)
        return dp[prev][count];
        
        long total=0;
        for(int i=prev;i<=N;i+=prev)
        {
            total+=find(i,count-1)%MOD;
            if(total>=MOD)
            total%=MOD;
        }
        return dp[prev][count]=total;
    }
   
}