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 C
{
	
	static int mod = (int)1e9+7;
	static int dp[][];
	
	static int add(int x,int y)
	{
		return (x%mod + y%mod)%mod;
	}
	
	static int sim(int n,int k,int x,int s)
	{
		if(x>n)
			return 0;
		if(s==k)
			return 1;
		
		if(dp[x][s]!=-1)
			return dp[x][s];
		
		int sum = 0;
		
		for(int i=1;i<=n/x;i++)
			sum = add(sum,sim(n,k,x*i,s+1));
		
		return dp[x][s] = sum;
		
	}

	public static void main(String[] args)
	{
		Scanner sc = new Scanner(System.in);
		
		int n = sc.nextInt();
		int k = sc.nextInt();
		
		dp = new int[n+1][k+1];
		for(int i=0;i<n+1;i++)
			for(int j=0;j<k+1;j++)
				dp[i][j] = -1;
		
		int sum = 0;
		
		for(int i=1;i<=n;i++)
		{
			sum=add(sum,sim(n,k,i,1));
		}
		
		System.out.println(sum);
		
	}
	
}