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:
# import sys
# sys.stdin = open("input.in","r")
# sys.stdout = open("output.out","w")

n,k1 = map(int,input().split())

dp = [[0 for i in range(n+1)] for j in range(k1+1)]
dp[0][0] = 1
for i in range(k1+1):

	dp[i][0] = 1

# for i in range(n+1):

# 	dp[i][0] = 1

for i in range(1,n+1):

	dp[1][i] = 1

# print(dp)a


for i in range(2,k1+1):

	for j in range(1,n+1):
		k = j
		# r = j
		# print("--")
		while k <= n:

			dp[i][k] = (dp[i][k] + dp[i-1][j])%1000000007

			k = k + j
# print(dp)
print((sum(dp[k1])-1)%1000000007)		



# print(dp)
# print(dp[n][k])