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 {

	private static final int MOD = 1000000007;

	public static void main(String[] args) {
		try (Scanner s = new Scanner(System.in)) {
			int upperBound = s.nextInt();
			int sequenceLength = s.nextInt();
			System.out.println(numSequences(upperBound, sequenceLength));
		}
	}

	private static int numSequences(int upperBound, int sequenceLength) {
		// Store the number of sequences starting with the first index, of length equal
		// to the second index, both zero indexed.
		int[][] startWithLength = new int[upperBound][sequenceLength];
		for (int i = 0; i < upperBound; i++) {
			// Length 1 can only have 1 sequence for each number.
			startWithLength[i][0] = 1;
		}
		for (int currentLength = 2; currentLength <= sequenceLength; currentLength++) {
			for (int currentStart = 1; currentStart <= upperBound; currentStart++) {
				int sum = 0;
				for (int multiplier = 1; currentStart * multiplier <= upperBound; multiplier++) {
					sum = (sum + startWithLength[currentStart * multiplier - 1][currentLength - 2]) % MOD;
				}
				startWithLength[currentStart - 1][currentLength - 1] = sum;
			}
		}
		// Add up the all combinations of the target length to get the final answer.
		int sum = 0;
		for (int i = 0; i < upperBound; i++) {
			sum = (sum + startWithLength[i][sequenceLength - 1]) % MOD;
		}
		return sum;
	}

}