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 lolat {
	static int n;

	static long[][] memo;

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		while (in.hasNext()) {
			n = in.nextInt();
			int k = in.nextInt();
			memo = new long[n + 1][k + 1];
			for (int i = 0; i <= n; i++) {
				Arrays.fill(memo[i], -1);
			}
			System.out
					.println(callRecusive(1, k) % 1000000007);
		}
	}

	private static long callRecusive(int num, int len) {
		// TODO Auto-generated method stub
		if (len == 0) {
			return 1;
		}
		if (memo[num][len] != -1)
			return memo[num][len];
		long s = 0;
		for (int i = num; i <= n; i += num) {
			s += callRecusive(i, len - 1);
		}
		return (memo[num][len] = s % 1000000007);
	}

}