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 math
n, k = [*map(int, input().split())]

prime = [1]
f = [False] * 2000
for i in range(2, 2000):
    if f[i]:
        continue
    f[i] = True
    prime.append(i)
    j = 1
    while i * j < 2000:
        f[i * j] = True
        j += 1

f = [[0] * (n + 1) if i != 1 else [1] * (n + 1) for i in range(k + 1)]
f[1][0] = 0
for i in range(1, n + 1):
    j = 1
    div = []
    while j <= i:
        if not (i % j):
            div.append(j)
        j += 1
    for j in range(2, k + 1):
        for l in div:
            f[j][i] += f[j - 1][i // l]
            f[j][i] %= 10 ** 9 + 7
ans = 0
for i in f[k]:
    ans += i
    ans %= 10 ** 9 + 7
print(ans)