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.*;
import java.math.*;
import java.io.*;

public class Main{
	static int n = 0;
	static long[][] dp = new long[3000][3000]; 
	static long MOD = (long)1e9 + 7L;

	public static void main(String[] args) throws FileNotFoundException{
		for(long[] a : dp)
			Arrays.fill(a, -1L);

		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		int k = sc.nextInt();
		System.out.println(count(1, k) % MOD);
		/*PrintWriter out = new PrintWriter(new File("out.txt"));
		n = 1;
		while(n <= 2000){
			for(int k = 1; k <= 2000; k++){
				long res = count(1, k) % MOD;
				out.println("res[" + n + "][" + k + "] = " + res + ";");
				out.flush();
			}
			n++;
		}
		out.close();*/
	}

	static long count(int last, int k){
		if(last > n)return 0L;
		if(k == 0)return 1L;	

		if(dp[last][k] != -1L){
			return dp[last][k];
		}

		long sum = 0L;
		for(int i = 1; i <= n; i++){
			if(last * i > n)
				break;
			sum += count(last * i, k - 1) % MOD;
		}

		return dp[last][k] = sum % MOD;
	}
}