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 Main {

    private static final int MOD = 1_000_000_007;

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int N = in.nextInt();
        int K = in.nextInt();

        int[][] dp = new int[K + 1][N + 1];
        // dp[len][nr] = number of good sequence of length len that ends in nr

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

        for (int len = 1; len < K; len++) {
            for (int nr = 1; nr <= N; nr++) {
                for (int mult = nr; mult <= N; mult += nr) {
                    dp[len + 1][mult] = (dp[len + 1][mult] + dp[len][nr]) % MOD;
                }
            }
        }

        int answer = 0;

        for (int nr = 0; nr <= N; nr++) {
            answer = (answer + dp[K][nr]) % MOD;
        }

        System.out.println(answer);
    }
}