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;

/**
 * Created by manan Sharma on 03-Aug-16.
 */
public class Solution {
	public static void main( String[] args ) throws IOException {
		int n, k;
		BufferedReader inputStream = new BufferedReader( new InputStreamReader( System.in ) );
		String input = inputStream.readLine();
		String[] inputs = input.split( " " );
		n = Integer.parseInt( inputs[ 0 ] );
		k = Integer.parseInt( inputs[ 1 ] );
		long[][] dp = new long[ n + 1 ][];
		boolean[] isPresent = new boolean[ dp.length ];
		for ( int i = 1; i <= n; ++i ) {

			for ( int j = i; j <= n; j += i ) {
				if ( !isPresent[ j ] ) {
					isPresent[ j ] = true;
					dp[ j ] = new long[ k + 1 ];
					dp[ j ][ 0 ] = 0;
					dp[ j ][ 1 ] = 1;
				}
				for ( int ij = 1; ij <= k; ++ij ) {
					dp[ j ][ ij ] = ( dp[ i ][ ij - 1 ] + dp[ j ][ ij ] ) % 1000000007;
				}
			}
		}
		long v = 0;
		for ( int i = 1; i <= n; ++i ) {
			v = ( v + dp[ i ][ k ] ) % 1000000007;
		}
		System.out.println( v );
	}
}