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 PYTHON):

Solution:
def cf(k, m):
    # num of solutions to
    # x_1 + x_2 + ... + x_m = k, x_i \in [0, 1] = C(k+m-1, k-1)
    # so \prod_i p_j^{x_j_i} = p_j^{e_j},
    # and \prod_j \prod_i p_j^{x_j_i} = n
    c = 1
    # Do C(k+m-1, m) since m < 11 (k < 2^11)
    for i in range(1, m+1):
        c *= k+m-i
        c //= i
    return c

N, K = map(int, raw_input().strip().split())
M = 1000000007
ans = 1
for n in range(2, N+1):
    cp = 1  # ways to end at n, n = \prod_j p_j^e_j
    t = n
    for j in range(2, n+1):
        ej = 0
        while t%j==0:
            t//= j
            ej += 1
        if ej > 0:
            cp *= cf(K, ej)
            cp %= M
    ans += cp
    ans %= M
print ans