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:
#Author: squiggly_lines
#Date: 30/04/2014
#Problem: 414B
import math
def main():
    n,k = map(int, raw_input().split())
    print B(n,k) % (10**9+7)
    
def B(n,k):
    s = 0
    for i in xrange(1,n+1):
        s += A(i,k)

    return s

def A(n,k): #Sometimes I forget that code doesn't have to be translated directly ...
    if n == 1:
        return 1
    pf = prime_factor(n)

    prod = 1
    for (p, exp) in pf:
        prod *= binom(exp+k-1, k-1)

    return prod

def prime_factor(n):
    pf = []
    d = 2
    while d**2 <= n:
        exp = 0
        while n % d == 0:
            n /= d 
            exp += 1
        if exp > 0:    
            pf.append((d,exp)) #The prime doesn't actually matter, but whatever
        d += 1
    if n > 1:
        pf.append((n,1))
    return pf

def binom(n,k): #Because importing itertools or scipy seems overkill
    return reduce(mult, xrange(k+1,n+1)) / reduce(mult, xrange(1,n-k+1))

def mult(x,y): #To compensate for not knowing Python
    return x*y

if __name__ == "__main__":
    main();