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.util.*;

public class mashmokh_and_acm
{
    public static void main( String[] args )
    {
        Scanner in = new Scanner( System.in );
        int n = in.nextInt(), k = in.nextInt();
        long[][] dp = new long[k][n + 1];
        for ( int i = 1; i <= n; i++ )
            dp[0][i] = 1;
        for ( int a = 1; a < k; a++ )
            for ( int b = 1; b <= n; b++ )
                for ( int c = 1; c <= n / b; c++ )
                {
                    dp[a][b * c] += dp[a - 1][b];
                    dp[a][b * c] %= 1000000007;
                }
        long sum = 0;
        for ( int i = 1; i <= n; i++ )
            sum += dp[k - 1][i];
        System.out.println( sum % 1000000007 );
        in.close();
        System.exit( 0 );
    }
}