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 {
    static int MOD = 1000000007;

    public static void main(String args[]) {
        Scanner scan = new Scanner(System.in);
        while (scan.hasNext()) {
            int n = scan.nextInt();
            int k = scan.nextInt();
            int dp[][] = new int[2005][2005];
            for (int i = 0; i <2005; i++)
                for (int j = 0; j <2005; j++)
                    dp[i][j] = 0;
            for (int i = 1; i <= n; i++)
                dp[1][i] = 1;
            for (int len = 1; len < k; len++) {
                for (int i = 1; i <= n; i++) {
                    for (int j = 1; j * i <= n; j++) {
                        dp[len + 1][j * i] += dp[len][i];
                        dp[len + 1][j * i] %= MOD;
                    }
                }
            }
            int ans = 0;
            for (int i = 1; i <= n; i++) {
                ans += dp[k][i];
                ans %= MOD;
            }
            System.out.println(ans);

        }
    }
}