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 ACM {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int n = Integer.parseInt(st.nextToken()), k = Integer.parseInt(st.nextToken());
		long[][] dp = new long[k+1][n+1];
		ArrayList<Integer>[] divisors = new ArrayList[n+1];
		for(int i = 0; i <= n; i++) divisors[i] = new ArrayList<Integer>();
		for(int i = 1; i <= n; i++) {
			int mult = 1;
			while(mult * i <= n) {
				divisors[mult * i].add(i);
				mult++;
			}
		}
		for(int i = 1; i <= k; i++) {
			for(int j = 1; j <= n; j++) {
				if(i == 1) {
					dp[i][j] = 1;
				} else {
					long sum = 0;
					for(Integer m: divisors[j]) {
						sum = (long)(((long)sum + dp[i-1][m]) % 1000000007) ;
					}
					dp[i][j] = sum;
				}
			}
		}
		long ret = 0;
		for(int i = 1; i <= n; i++) {
			ret = (ret + dp[k][i]) % 1000000007;
		}
		System.out.println(ret);
		//for(int i = 1; i <= k; i++) System.out.println(Arrays.toString(dp[i]));
	}
}