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:
'''
	CodeForces 414B
	Mashmokh and ACM

	Tags: dp, Math
'''

MOD = 1000000007

n, k = tuple(map(int, input().split()))

# divider = [ [ x for x in range(1, n + 1) if i % x == 0 ] for i in range(n + 1) ]

dp = [ [ 1 if j == 1 else 0 for j in range(k + 1) ] for i in range(n + 1) ]

for j in range(1, k):
	for i in range(1, n + 1):
		for ii in range(i, n + 1, i):
			dp[ii][j + 1] = (dp[ii][j + 1] + dp[i][j]) % MOD

ans = sum([ dp[i][k] for i in range(1, n + 1) ]) % MOD
print(ans)