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.Arrays;

public class MokhSeries {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));

        String[] input = br.readLine().trim().split(" ");
        int n = Integer.parseInt(input[0]);
        int k = Integer.parseInt(input[1]);

        int[] dpA = new int[n + 1];
        int[] dpB = new int[n + 1];

        Arrays.fill(dpA, 1);
        dpA[0] = 0;

        for (int i = 1; i < k; i++) {
            for (int j = 1; j < n + 1; j++) {
                for (int index = 1; index < dpA.length && index * j < n + 1; index++) {
                    dpB[index * j] += dpA[index];
                    dpB[index * j] %= 1000000007;
                }
            }
            for (int j = 0; j < n + 1; j++) {
                dpA[j] = dpB[j];
                dpB[j] = 0;
            }
        }
        int sum = 0;
        for (int num : dpA) {
            sum += num;
            sum %= 1000000007;
        }

        bw.write(String.format("%d%n", sum));

        br.close();
        bw.close();
    }
}