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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class D240Again {
	static int[][] dp;
	static int mod = 1000000007;
	static int n; 
	static int getValue(int length, int last) {
		if (length == 1) {
			return 1;
		}
		if (dp[length][last] != -1) {
			return dp[length][last];
		}
		int value = 0;
		for (int i = 1; i <= (n/last); i++) {
				value = (value + getValue(length - 1, last* i)) % mod;
		}
		return dp[length][last]  =value;
	}

	public static void main(String[] args) throws IOException {
		
		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
		String[] line = bf.readLine().split(" ");
		n = Integer.parseInt(line[0]);
		int k = Integer.parseInt(line[1]);
		dp = new int[k+2][n+2];
		for (int i = 0; i <= k+1; i++) {
			for (int j = 0; j <= n+1; j++) {
				dp[i][j] = -1;
			}
		}
		int value = 0;
		for (int i = 1; i <= n; i++) {
			value = (value + getValue(k, i)) % mod;
		}
		System.out.println(value);
	}
}