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.Arrays;
import java.util.Scanner;

/**
 * Created by fly on 9/4/17.
 */
public class Main {
    public static int mod = 1000000007;

    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];
        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) {
                int s = 1;
                int mul = j;
                do {
                    dp[i + 1][mul] += dp[i][j];
                    dp[i+1][mul] %= mod;
                    ++s;
                    mul = s * j;
                } while(mul <= n);
            }
        }
        int sum = 0;
        for (int i = 1; i <= n; ++i) {
            sum += dp[k][i];
            sum %= mod;
        }
        System.out.println(sum);
    }
}