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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class D {

  final int mod = 1000000007;

  private Map<Integer, List<Integer>> cache = new HashMap<Integer, List<Integer>>();

  private List<Integer> div(int j) {
    if (!cache.containsKey(j)) {
      List<Integer> res = new ArrayList<Integer>();
      for (int i = 1; i <= j; ++i) {
        if (j % i == 0)
          res.add(j / i);
      }
      cache.put(j, res);
    }
    return cache.get(j);
  }

  private void solve() throws IOException {
    int n =ni();
    int k = ni();

    int m[][] = new int[n + 1][k + 1];

    for (int i = 1; i <= n; ++i) {
      m[i][1] = 1;
    }

    for (int i = 2; i <= k; ++i) {
      for (int j = 1; j <= n; ++j) {
        for (Integer f : div(j)) {
          m[j][i] = (m[j][i] + m[f][i - 1]) % mod;
        }
      }
    }

//    for (int i = 1; i <= k; ++i) {
//      for (int j = 1; j <= n; ++j) {
//        pr(m[j][i] + "\t");
//      }
//      prln();
//    }

    int res = 0;
    for (int i = 1; i <= n; ++i) {
      res = (res + m[i][k]) % mod;
    }
    prln(res);
  }

  public static void main(String ... args) throws IOException { new D().run(); }

  private PrintWriter pw;
  private BufferedReader br;
  private StringTokenizer st;

  private void run() throws IOException {
    pw = new PrintWriter(System.out);
    br = new BufferedReader(new InputStreamReader(System.in));
    solve();
    br.close();
    pw.close();
  }

  private void pr(Object o) { pw.print(o); }
  private void prln(Object o) {pw.println(o);}
  private void prln() {pw.println();}

  private long nl() throws IOException { return Long.parseLong(nt());	}

  private double nd() throws IOException  { return Double.parseDouble(nt()); }

  private int ni()throws IOException  { return Integer.parseInt(nt()); }

  private String nt() throws IOException {
    if (st == null || !st.hasMoreTokens())
      st = new StringTokenizer(br.readLine());
    return st.nextToken();
  }
}