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:


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
# for row in dp:
# 	print(row)
print(s)
#  C:\Users\Usuario\HOME2\Programacion\ACM