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.lang.*;
import java.io.*;
import java.util.*;


public class Main {

  public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    PrintWriter out = new PrintWriter(System.out, true);
    new Task().solve(in, out);
    out.close();
    in.close();
  }
}

class Task {
  public void solve(Scanner in, PrintWriter out) {
    int n = in.nextInt();
    int k = in.nextInt();
    int[][] dp = new int[k+1][n+1];
    for (int i=0; i<=n; i++) {
      dp[1][i] = 1;
    }
    for (int i=2; i <= k; i++) {
      for (int j=1; j <= n; j++) {
        for(int h=1; j * h <= n; h++) {
          dp[i][j * h] = (dp[i][j * h] + dp[i-1][j]) % 1000000007;
        }
      }
    }
   // for (int i=0; i <= k; i++) {
   //   for (int j=0; j <= n; j++) {
   //     out.print(dp[i][j]);
   //     out.print(" ");
   //   }
   //   out.println();
   // }
    int result = 0;
    for (int j=1; j <= n; j++) {
      result = (result + dp[k][j]) % 1000000007;
    }
    out.println(result);
  }
}