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 Main {

    public static void main(String[] args) {

        Scanner in = new Scanner(System.in);

        int n = in.nextInt();
        int k = in.nextInt();

        int[][] divisor = new int[2014][45];

        divisor[1][0] = 1;
        divisor[1][1] = 1;

        for (int i = 2; i <= n; i++) {

            for (int j = 1; j <= i; j++) {

                if (i % j == 0) {

                    divisor[i][0]++;
                    divisor[i][divisor[i][0]] = j;
                }
            }
        }

        int[][] res = new int[2014][2014];

        for (int i = 1; i <= n; i++)
            res[1][i] = 1;

        for (int i = 2; i <= k; i++) {

            for (int j = 1; j <= n; j++)
                for (int q = 1; q <= divisor[j][0]; q++)
                    res[i][j] = (res[i][j] + res[i - 1][divisor[j][q]]) % 1000000007;
        }

        long ans = 0;
        
        for (int i = 1; i <= n; i++)
            ans = (ans + res[k][i]) % 1000000007;

        System.out.print(ans);
    }

}