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 CONTEST {
  static final int MX = 2005;
  static final int mod = 1000000007;
  static int[][] ff = new int[MX][MX];
  
  static int dp( int n, int k ) {
    for(int i=0; i<=k; i++) {
      for(int j=0; j<=n; j++) {
    	if( i==k ) ff[i][j] = 1;
    	else ff[i][j] = 0;
      }
    }
    
    for(int i=k-1; i>=0; i--) {
      for(int j=n; j>=1; j--) {
    	for(int m=j; m<=n; m+=j) {
    	  ff[i][j] = (ff[i][j]%mod + ff[i+1][m]%mod)%mod;
    	}
      }
    }
    
    return (ff[0][1]%mod);
  }
  
  public static void main( String args[] ) {
    Scanner sc = new Scanner( System.in );
    PrintWriter pf = new PrintWriter( System.out );
    int n, k;
    n = sc.nextInt();
    k = sc.nextInt();
    int ans = dp( n, k ) % mod; 
    pf.println(ans);
    
    pf.close();
  }
}