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;
import java.io.PrintWriter;
import java.util.*;


public class B {	
	BufferedReader reader;
    StringTokenizer tokenizer;
    PrintWriter out;
    
	public void solve() throws IOException {
		int K = nextInt();
		int N = nextInt();		
		
		long[] dp = new long[K+1];
		Arrays.fill(dp, 1);
		dp[0] = 0;
		
		for (int i = 1; i < N; i++) {
			long[] old = dp;
			dp = new long[K+1];
			for (int j = 1; j <= K; j++) {
				int k = j;
				while (k <= K) {
					dp[k] = mod_add(dp[k], old[j]);
					k += j;
				}
			}
		}
		
		long ans = 0;
		for (int i = 1; i <= K; i++) {
//			out.println(i + ": " + dp[i]);
			ans = mod_add(ans, dp[i]);
		}
		out.println(ans);
		
	}
	
	long MOD = (long) 1e9 + 7;
	public long mod_add(long n1, long n2){
    	return (n1 + n2) % MOD;
    }    
	
	/**
	 * @param args
	 */
	public static void main(String[] args) {
		new B().run();
	}
	
	public void run() {
        try {
            reader = new BufferedReader(new InputStreamReader(System.in));
            tokenizer = null;
            out = new PrintWriter(System.out);
            solve();
            reader.close();
            out.close();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(1);
        }
    }

    int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    String nextToken() throws IOException {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(reader.readLine());
        }
        return tokenizer.nextToken();
    }

}