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:
//package codeforces;

import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;


public class MashmokhandACM {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        
        int m = 1000000007;
        int n = in.nextInt(),
            k = in.nextInt();
            
        if (k == 1) {
            out.println(n);
        }
        else {
            long[] dp_prev = new long[n+1];
            Arrays.fill(dp_prev, 1);
            dp_prev[0] = 0;
        
            long[] dp = new long[n+1];
            for (int i = 1; i < k; i++) {
                for (int j = 1; j <= n; j++) {
                    long sum = 0;
                    for (int num = j; num <= n; num += j)
                        sum = (sum + dp_prev[num]) % m;                    
                    
                    dp[j] = sum;
                }
                
                System.arraycopy(dp, 1, dp_prev, 1, n);
            }
            
            long sum = 0;
            for (long i : dp) sum = (sum + i) % m;
            out.println(sum);
        }
        
        out.flush();
    }
}