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.math.BigInteger;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Main {
	static final int MOD_DIVISOR = 1_000_000_007;

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		int n = sc.nextInt();
		int k = sc.nextInt();
		System.out.println(solve(n, k));

		sc.close();
	}

	static int solve(int n, int k) {
		int result = 0;

		Map<Integer, Integer> lastToWayNum = new HashMap<>();
		BigInteger c = BigInteger.ONE;
		for (int i = 0; i < k; i++) {
			Map<Integer, Integer> nextLastToWayNum = new HashMap<>();

			if (i == 0) {
				for (int j = 1; j <= n; j++) {
					nextLastToWayNum.put(j, 1);
				}
			} else {
				for (int last : lastToWayNum.keySet()) {
					for (int nextLast = last * 2; nextLast <= n; nextLast += last) {
						nextLastToWayNum.put(nextLast,
								addMod(nextLastToWayNum.getOrDefault(nextLast, 0), lastToWayNum.get(last)));
					}
				}
			}

			if (i != 0) {
				c = c.multiply(BigInteger.valueOf(k - i)).divide(BigInteger.valueOf(i));
			}

			result = addMod(result, multiplyMod(nextLastToWayNum.values().stream().reduce(0, Main::addMod),
					c.mod(BigInteger.valueOf(MOD_DIVISOR)).intValue()));

			lastToWayNum = nextLastToWayNum;
		}

		return result;
	}

	static int addMod(int x, int y) {
		return (x + y) % MOD_DIVISOR;
	}

	static int multiplyMod(int x, int y) {
		return (int) ((long) x * y % MOD_DIVISOR);
	}
}