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:


////////////////////---------------------------SHUBHAM CHAUDHARI-------------------------------///////////////////////


import java.util.*;

public class Main {

    static long dp[][]=new long[101][2];
    static long m=1000000007;
    static ArrayList<Integer>graph[];
    public static void main(String[] args){
        Scanner in=new Scanner(System.in);

        long mod=1000000007;
        int n=in.nextInt();
        int k=in.nextInt();
        ArrayList<Integer>num[]=new ArrayList[n+1];
        for(int i=0;i<=n;i++)
        {
            num[i]=new ArrayList<>();
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=i;j++)
            {
                if(i%j==0)
                {
                    num[i].add(j);
                }
            }
        }
        long dp[][]=new long[k+1][n+1];
        for(int i=1;i<n+1;i++)
        {
            dp[1][i]=1;
        }
        for(int i=2;i<=k;i++)
        {
            for(int j=1;j<=n;j++)
            {
                for(int l=0;l<num[j].size();l++)
                {
                    dp[i][j]=(dp[i][j]+dp[i-1][num[j].get(l)]+mod)%mod;
                }
            }
        }
        long res=0;
        for(int i=1;i<=n;i++)
        {
            res=(res+dp[k][i]+mod)%mod;
        }
        System.out.println((res+mod)%mod);
    }
//n>1


}