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.util.Scanner;

public class MashmokhAndAcm {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt(), k = sc.nextInt();
        long s = 0, mod = 1000000007;
        long[] dp = new long[n+1];
        if(k == 1) {
            System.out.println(n);
            return;
        }
        for(int i = 1; i < n+1; i++){
            dp[i] = (int) Math.floor((n * 1.0 / i)) % mod;
            s += dp[i];
        }
        for(int i = 1; i < k-1; i++){
            for(int j = 2; j <= (n/2); j++){
                for(int l = 2*j; l <= n; l += j){
                    dp[j] += dp[l];
                    dp[j] %= mod;
                }
                s += dp[j];
                s %= mod;
            }
            s += (n-(n/2))%mod;
        }
        System.out.println(s % mod);
    }
}