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.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
import java.util.Scanner;

public class JJ {

	
	static long dp[][];
	static int len;
	static int n;
	
	public static void main(String[] args) {
		
		Scanner in = new Scanner(System.in);
		
		 n = in.nextInt();
		
		 len = in.nextInt();
		 
		 dp = new long[n+1][len+1];
		 
		 for(int i=0; i<dp.length; i++)
			 for(int j=0; j<dp[i].length; j++) {
				 dp[i][j] = -1;
			 }
		
		 
		long ans = 0;
		for(int i=1; i<=n; i++) {
			ans += rec(i, len-1) ;
			ans %= 1000000007;
		}
		
		System.out.println(ans);
		
	}

	private static long rec(int i, int len) {
		if(len == 0)
			return 1;
		else if(dp[i][len] != -1)
			return dp[i][len];
		else {
			long ans = 0;
			for(int k=i; k <= n; k+=i) {
				if(k % i == 0) {
					ans += rec(k, len-1) ;
					ans %= 1000000007;
				}
			}
			
			return dp[i][len] = ans ;
		}
	}
}