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 PYTHON3):

Solution:
# Description of the problem can be found at http://codeforces.com/problemset/problem/414/B

n,k = map(int,input().split())
n += 1
k += 1
dp = [[0] * n for _ in range(13)]
s = 0
M = 1000000007
f = [1]
for i in range(1,4500):
	f.append(f[-1] * i % M)
def C(n, k):
	if k>n:
		return 0
	return  f[n] * pow(f[k] * f[n - k] %M , M - 2, M) % M

for i in range(1, 12):
	for j in range(1, n):
		if i == 1:
			dp[i][j] = 1
		for t in range(j * 2, n, j):
			dp[i + 1][t] = (dp[i + 1][t] + dp[i][j]) % M
		s = (s + C(k - 2, i - 1) * dp[i][j] % M) % M


print(s)