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 ACM {
 
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt(), k = scan.nextInt(), mod = 1000000007;
		long[][] dp = new long[k][n+1];
		for(int i = 0; i < k; i++) for(int j = 1; j <= n; j++){
			if(i == 0) dp[i][j] = 1;
			if(i < k-1) for(int l = 1; l*j <= n; l++){
				dp[i+1][j*l] += dp[i][j];
				dp[i+1][j*l] %= mod;
			}
		}
		long sum = 0;
		for(int i = 1; i <= n; i++) sum = (sum+dp[k-1][i])%mod;	
		System.out.println(sum);
	}
 
}