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.*;
import java.io.*;
public class A
{
      public static void main(String ar[]) throws Exception
      {
            BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
            String s1[]=br.readLine().split(" ");
            int n=Integer.parseInt(s1[0]);
            int k=Integer.parseInt(s1[1]);
            long mod=1000000007;
            long dp[][]=new long[n+1][k+1];
            for(int i=1;i<=n;i++)
             dp[i][1]=1;
           
            ArrayList<Integer> al[]=new ArrayList[n+1];
            for(int i=1;i<=n;i++)
            {
                  al[i]=new ArrayList<Integer>();
                  for(int j=1;j*j<=i;j++)
                  {
                        if(i%j==0)
                        { al[i].add(j);   
                          if(j!=i/j)
                         al[i].add(i/j); 
                        }
                  }
            }
            
            for(int i=2;i<=k;i++)
            {
                  for(int j=1;j<=n;j++)
                  {
                        for(int p=0;p<al[j].size();p++)
                        { int r=(int)al[j].get(p); dp[j][i]=(dp[j][i]+dp[j/r][i-1])%mod; }
                  }
            }
            
            long S=0;
            for(int i=1;i<=n;i++)
            {
                  //System.out.println(al[i]);
                  S=(S+dp[i][k])%mod;
            }
            System.out.println(S);
      }
}