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.Arrays;
import java.util.Scanner;

public class MashmokhandACM {
	static int n, k;
	static long memo[][];

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		k = sc.nextInt();

		memo = new long[k + 1][n + 1];
		for (int i = 0; i < k + 1; i++)
			Arrays.fill(memo[i], -1);

		System.out.println((int) (dp(0, 1) % ((1e9) + 7)));

	}

	static long dp(int pos, int val) {

		if (pos == k)
			return 1;
		if (memo[pos][val] != -1)
			return memo[pos][val];

		long ans = 0;
		for (int i = val; i <= n; i += val)
			ans += (int)(dp(pos + 1, i) % ((1e9) + 7));

		return memo[pos][val] = ans;

	}
}