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.Scanner;

public class D{
    
    static int MOD = 1000000007;
    
    public static void main(String[] args){
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int k = in.nextInt();
        int dp[][] = new int[k + 1][n + 1];
        for(int i = 0 ; i <= k ; i++)
            for(int j = 0 ; j <= n ; j++)
                dp[i][j] = -1;
        int ans = 0;
        for(int i = 1 ; i <= n ; i++)
            ans = (ans + solve(k - 1,i,n,dp)) % MOD;
        System.out.println(ans);             
    }    
    
    static int solve(int sisa,int mul,int lmt,int dp[][]){
        if(sisa == 0)
            return 1;
        if(dp[sisa][mul] != -1)
            return dp[sisa][mul];
        dp[sisa][mul] = 0;
        for(int i = mul ; i <= lmt ; i += mul)
            dp[sisa][mul] = (dp[sisa][mul] + solve(sisa - 1,i,lmt,dp)) % MOD;
        return dp[sisa][mul];            
    }
    
}