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:
// package Dynamic;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class MashMokh {
    public static void print(long dp[][]){
        for(int i=1;i<dp.length;i++){
            for(int j=1;j<dp[0].length;j++){
                System.out.print(dp[i][j]+" ");
            }
            System.out.println();
        }
    }
    public static void main(String[] args)throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        String line[]=br.readLine().split(" ");
        int n=Integer.parseInt(line[0]);
        int k=Integer.parseInt(line[1]);
        long dp[][]=new long[n+1][k+1];
        int mod=1000000007;
        for(int i=n;i>=1;i--){
            dp[i][1]=1;
            for(int j=2;j<=k;j++){
                dp[i][j]=dp[i][j-1];
                for(int row=2*i;row<=n;row+=i){
                    dp[i][j]=(dp[i][j]+dp[row][j-1])%mod;
                }
            }
        }
//        print(dp);
        long ans=0;
        for(int i=1;i<=n;i++){
            ans=(ans+dp[i][k])%mod;
        }
        System.out.println(ans);
    }
}