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.ArrayList;
import java.util.List;
import java.util.Scanner;

/**
 * Created by rudolf on 06.04.2014.
 */
public class TaskD {
    public static void main(String[] args) {
        int MOD = 1000000007;
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        long st = System.currentTimeMillis();
        Integer divisors[][] = new Integer[n+1][];
        for(int i = 1;i <=n;i++) {
            List<Integer> tmp = new ArrayList<Integer>();
            for(int j = 1; j<=i;j++)
                if(i % j == 0)
                    tmp.add(j);
            divisors[i] = new Integer[tmp.size()];
            tmp.toArray(divisors[i]);
        }
        long dp[][] = new long[k + 1][n + 1];
        dp[0][1] = 1;
        for (int i = 1; i < dp.length; i++)
            for (int j = 1; j < dp[i].length; j++) {
                dp[i][j] = 0;
                for(int c = 0; c < divisors[j].length; c++)
                    dp[i][j] += dp[i-1][divisors[j][c]];
//                for (int c = 1; c <= j; c++) {
//                    if(j % c == 0) dp[i][j] += dp[i-1][c];
//                }
                dp[i][j] %= MOD;
            }
        long res = 0;
        for(int i = 0; i < dp[0].length; i++)
            res = (res + dp[dp.length - 1][i]) % MOD;
//        System.out.println("Time: " + (System.currentTimeMillis() - st));
        System.out.println(res);
    }
}