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.*;

public class DP 
{
    public static void main(String[] args)
    {
        int MOD = 1000000007;
        Scanner input = new Scanner(System.in); 
        int n = input.nextInt(); 
        int k = input.nextInt(); 
        int dp[][] = new int[2005][2005]; 
        
        for (int i = 1; i <= 2000; i++)
		    dp[1][i] = 1;
	
	    for (int i = 1; i <= (k-1); i++)
		    for (int j = 1; j <= n; j++)
		    	for (int l = 1; l*j <= n; l++)
		    {
				    dp[i+1][l*j] += dp[i][j];
				    dp[i+1][l*j] -= dp[i+1][l*j]/MOD*MOD;
			}
	
	int result = 0;
	for (int i = 1; i <= n; i++) 
	{
		result += dp[k][i];
		result -= result/MOD*MOD;
	}
            
        System.out.println(result);
    }
}