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:
//package com.himanshu.practice.july.july23;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

/**
 * Created by himanshubhardwaj on 23/07/18.
 */
public class MashmokhAndACM {

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n, length;
        String st[] = br.readLine().split(" ");
        n = Integer.parseInt(st[0]);
        length = Integer.parseInt(st[1]);

        long DP[][] = new long[n + 1][length + 1];

        //DP[i][j] --> subsequence of length j ending at i
        for (int i = 1; i <= n; i++) {
            DP[i][1] = 1;
        }

        long sum = 0;

        for (int i = 2; i <= n; i++) {
            for (int j = 1; j < length; j++) {
                for (int k = 1; (k * i) <= n; k++) {
                    DP[k * i][j + 1] = (DP[k * i][j + 1] + DP[i][j]) % 1000000007;
                }
            }
        }

        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= length; j++) {
                sum = (sum + DP[i][j]) % 1000000007;
            }
        }
        System.out.print(sum);
    }
}