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.math.*;
import java.io.*;
import java.util.*;
public class C414B{
   public static final int MOD=1000000007;
   public static void main(String[] args) throws IOException{
      Scanner sc=new Scanner(System.in);
      int N=sc.nextInt(); int K=sc.nextInt();
      int[][] dp=new int[N+1][K+1];
      for(int x=0;x<N+1;x++)
         dp[x][1]=1;
      for(int x=1;x<K;x++)
         for(int y=1;y<N+1;y++)
            for(int z=1;z*y<=N;z++){
                  dp[z*y][x+1]=(dp[z*y][x+1]+dp[y][x])%MOD;
                  if(dp[z*y][x+1]<0) dp[z*y][x+1]+=MOD;}   
      int sum=0;
      for(int x=1;x<N+1;x++)
         sum=(sum+dp[x][K])%MOD;
      if(sum<0) sum+=MOD;
      System.out.println(sum);
   }
}