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.*;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
import static java.lang.Math.*;
import java.math.*;

public class acm {
    static int n;
    static long[][] memo;
    public static void main(String[] args) throws IOException {
        BufferedReader gi = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter go = new PrintWriter(System.out);
        int[] line = parseArray(gi);
        n = line[0];
        int k = line[1];
        memo = new long[n+1][k+1];
        go.println(dp(1,k));
        go.close();
    }
    
    static int[] parseArray(BufferedReader gi) throws IOException{
        String[] line = gi.readLine().split(" ");
        int[] ans = new int[line.length];
        for (int i = 0; i < line.length; i++) {
            ans[i] = Integer.parseInt(line[i]);
        }
        return ans;
    }
    
    static long naive(int multiple,int len){
       
        if ( len == 1){
            return n/multiple;
        }
        long ans = 0;
        int k = 1;
        while (multiple*k <= n){
            ans += naive(multiple*k, len-1);
            k++;
        }
        return ans;
    }
    static long dp(int multiple,int len){
       
        if ( len == 1){
            return n/multiple;
        }
         if ( memo[multiple][len]>0){
            return memo[multiple][len];
        }
        long ans = 0;
        int k = 1;
        while (multiple*k <= n){
            ans += dp(multiple*k, len-1);
            ans = ans%1000000007;
            k++;
        }
        memo[multiple][len] = ans;
        return ans;
    }
    
}