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.*;

public class MashMokh {
    static BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter printWriter = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

    public static void main(String[] args) throws IOException{
        String[] inputString = bufferedReader.readLine().split(" ");
        int n = Integer.parseInt(inputString[0]);
        int k = Integer.parseInt(inputString[1]);
//        int n = 3; int k = 2;
        // given two integers n and k, find how many sequences with length k have such pattern:
        // each number divides by the next one without remainder and the sequence is non-decreasing

        // first of all, the sequence with the same number satisfies the requirement

        // use dp to solve the probelm
        // dp[i,k] means, current largest value is i, and its length is k, and the value is the number of such sequence
        // dp[i,k] = di[p,k-1] with all p < i, and p % i == 0
        // then the result is all  dp[i,k], i from 1 to n
        long modular = 1000000007;
        long[][] dp = new long[k+1][n+1];
        for(int i = 1; i <= n; i++) {
            dp[1][i] = 1;
        }

        for(int i = 2; i <= k; i++) {
            for(int j = 1; j <= n; j++) {
                    long base = 0;



                int range = (int)Math.sqrt(j);



                for(int t = 1; t <= range; t++) {

                    int p1 = j / t;
                    if( p1*t == j) {
                        base += dp[i-1][t];
                        if(p1 != t) base += dp[i-1][p1];
                    }
                }




                dp[i][j] = base % modular;
            }
        }

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

        ans %= modular;

        printWriter.println(ans);
        printWriter.flush();
        printWriter.close();



    }
}