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.*;
import java.util.*;

public class c1 {
    public static FScanner scan;
    public static PrintWriter out;
    public static void main(String[] args) {
        scan=new FScanner();
        out=new PrintWriter(new BufferedOutputStream(System.out));

        int T=1;
//        int T=scan.nextInt();
        while(T-->0) {

            long mod=(long)(1e9+7);
            int n=scan.nextInt(),k=scan.nextInt(),ans=0;
            int[][] dp=new int[n+1][k];
            for(int i=1;i<=n;i++) dp[i][0]=1;

            for(int i=1;i<k;i++) {
                for(int j=1;j<=n;j++) {
                    for(int m=1;m<=n;m++) {
                        if(j*m>n) break;
                        dp[j*m][i]+=dp[j][i-1];
                        dp[j*m][i]%=mod;
                    }
                }
            } for(int i=1;i<=n;i++) {
                ans+=dp[i][k-1];
                ans%=mod;
            } out.println(ans);

        } out.close();
    }



    //------------------------------------------------------------------------------------------------------------------

    public static class FScanner {
        BufferedReader br;
        StringTokenizer st;

        public FScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}