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.FileNotFoundException;
import java.util.Scanner;

import static java.lang.System.out;

public class Main {

    Scanner scanner = new Scanner(System.in);

    public static void main(String[] args) throws FileNotFoundException {
        new Main().solve();
    }

    void solve() throws FileNotFoundException {
        //Scanner scanner = new Scanner(new File("input.txt"));
        //PrintWriter out = new PrintWriter(new File("output.txt"));

        int n = scanner.nextInt(), k = scanner.nextInt();
        int z [] = new int[n + 1];

        /*
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= i; j++) {
                z[j] += i % j == 0 ? 1: 0;
            }
        }
        */

        long dp[][] = new long[k + 1][n + 1];

        for (int i = 1; i <= n; i++) {
            dp[1][i] = 1;
        }

        for (int i = 1; i < k; i++) {
            for (int j = 1; j <= n; j++) {
                for (int l = 1;  j * l <= n; l++) {
                    dp[i + 1][l * j] = (dp[i][j] + dp[i + 1][l * j]) % 1000000007;
                }
            }
        }

        long ans = 0;
        for (int i = 1; i <= n; i++) {
            ans = (ans + dp[k][i]) % 1000000007 ;
        }
        out.println(ans);
    }
}