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.util.ArrayList;
import java.util.StringTokenizer;

public final class MashmokhAndACM {
	public static void main(String args[]) throws IOException {
		BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer nk = new StringTokenizer(input.readLine());
		int n = Integer.parseInt(nk.nextToken());
		int k = Integer.parseInt(nk.nextToken());
		ArrayList<Integer> divisors[] = new ArrayList[n + 1];
		for (int i = 0; i <= n; i += 1)
			divisors[i] = new ArrayList<>();
		for (int i = 1; i <= n; i += 1) {
			for (int j = i; j <= n; j += i) {
				divisors[j].add(i);
			}
		}
		int dp[][] = new int[k][n + 1];
		for (int i = 0; i <= n; i += 1) {
			dp[0][i] = 1;
		}
		int mod = 1000000000 + 7;
		for (int len = 1; len < k; len += 1) {
			for (int end = 1; end <= n; end += 1) {
				for (int divisor : divisors[end]) {
					dp[len][end] = (dp[len - 1][divisor] + dp[len][end]) % mod;
				}
			}
		}
		int ans = 0;
		for (int i = 1; i <= n; i += 1) {
			ans = (ans + dp[k - 1][i]) % mod;
		}
		System.out.println(ans);
	}
}