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.awt.geom.Line2D;
    import java.math.BigInteger;
    import java.util.*;

    public class D {
        public static void main(String[] args) {
            long now = System.currentTimeMillis();
            Scanner in = new Scanner(System.in);
            int n = in.nextInt();
            int k = in.nextInt();
            int mod = 1000000007;
            int[][] dp =  new int[k + 1][n + 1];
            for (int i = 1; i <= n; i++) {
                dp[1][i] = 1;
            }
            List<Integer>[] mods = new List[n + 1];
            for (int i = 1; i <= n; i++) {
                mods[i] = new ArrayList();
                for (int j = 1; j <= i; j++) {
                    if (i % j == 0)
                        mods[i].add(j);
                }
            }
            for (int j = 2; j <= k; j++) {
                for (int i = 1; i <= n; i++) {
                    dp[j][i] = 0;
                    for (Integer integer : mods[i]) {
                        dp[j][i] = (dp[j][i] + dp[j-1][integer]) % mod;
                    }
                }
            }

            int result = 0;
            for (int i = 1; i <= n; i++) {
                result = (result + dp[k][i]) % mod;
            }
            System.out.println(result);
            in.close();
        }
    }