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.io.*;
import java.util.*;

public class A {
	public static void main(String[] args) throws IOException {
		Scanner s = new Scanner(System.in);
		BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		
		int MOD = 1000000007;
		int n = s.nextInt(), k = s.nextInt();
//		long start = System.currentTimeMillis();
		long[][] dp = new long[k][n+1];
		Arrays.fill(dp[0], 1);
		dp[0][0] = 0;
		for(int i = 1; i < k; i++) {
			for(int j = 1; j <= n; j++) {
				for(int l = j; l <= n; l+=j) {
					dp[i][j] += dp[i-1][l];
					dp[i][j] %= MOD;
				}
			}
		}
		long ans = 0;
		for(int i = 0; i <= n; i++) {
			ans += dp[k-1][i];
			ans %= MOD;
		}
		System.out.println(ans);
//		System.out.println(System.currentTimeMillis()-start);
	}
}