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.Scanner;

public class p414b {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

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

        scanner.close();

        memo = new int[n + 1][k + 1];
        System.out.println(f(1, k, n));
    }

    private static int[][] memo;

    private static int f(int last, int length, int max) {
        if (last > max) {
            return 0;
        }

        if (length == 0) {
            return 1;
        }

        if (memo[last][length] != 0) {
            return memo[last][length];
        }

        long count = f(last, length - 1, max);
        for (int i = last; i <= max; i += last) { // for each multiple of the last value in the seq
            count += f(last + i, length - 1, max);
            count %= 1000000007;
        }

        return memo[last][length] = (int) count;
    }
}