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 D {
	static void dbg(Object...os) {
		System.err.println(Arrays.deepToString(os));
	}
	static BufferedReader input;
	static StringTokenizer _stk;
	static String readln() throws IOException {
		String l = input.readLine();
		if(l!=null) _stk = new StringTokenizer(l," ");
		return l;
	}
	static String next() { return _stk.nextToken(); };
	static int nextInt() { return Integer.parseInt(next()); }
	static final int MOD=1000000000+7;
	public static void main(String[] args) throws IOException {
		input = new BufferedReader(new InputStreamReader(System.in));
		readln();
		int N=nextInt(), K = nextInt();
		long cur[]=new long[N+1];
		long next[]=new long[N+1];
		Arrays.fill(cur,1);
		for(int i=2; i<=K; i++) {
			Arrays.fill(next, 0);
			for(int j=1; j<=N; j++) {
				for(int k=j; k<=N; k+=j)
					next[k] = (next[k]+cur[j])%MOD;
			}

			long tmp[]=next;
			next=cur;
			cur=tmp;
		}
		long tot=0;
		for(int i=1; i<=N; i++) tot = (tot+cur[i])%MOD;

		System.out.println(tot);
	}


}