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:
//package codeforces;

import java.util.Scanner;

/**
 * Created by nitin.s on 15/01/17.
 */
public class MashmokhAndACM {
    static int mod = 1000000000 + 7;
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int k = in.nextInt();
        int[][] ans = new int[k + 1][n + 1];
        for(int i = 1; i <= n; ++i) {
            ans[1][i] = 1;
        }
        for(int i = 1; i < k; ++i) {
            for(int j = 1; j <= n; ++j) {
                for(int p = 1; p * j <= n; ++p) {
                    int s = p * j;
                    ans[i + 1][s] = (ans[i + 1][s] + ans[i][j]) % mod;
                }
            }
        }
        int sum = 0;
        for(int i = 1; i <= n; ++i) {
            sum = (sum + ans[k][i]) % mod;
        }
        System.out.println(sum);
    }
}