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:

from sys import maxsize, stdout, stdin,stderr
mod = int(1e9 + 7)

def I(): return int(stdin.readline())
def lint(): return [int(x) for x in stdin.readline().split()]
def S(): return input().strip()
def grid(r, c): return [lint() for i in range(r)]
from collections import defaultdict, Counter
import math
import heapq
from heapq import heappop , heappush
import bisect
from itertools import groupby
def gcd(a,b): 
    while b:
        a %= b
        tmp = a
        a = b
        b = tmp
    
    return a

def lcm(a,b): 
    return a  / gcd(a, b) * b

def check_prime(n):
    for i in range(2, int(n ** (1 / 2)) + 1):
        if not n % i:
            return False
    return True
def Bs(a, x):
    i=0
    j=0
    left = 0
    right = len(a)
    flag=False
    while left<right:
        
        mi = (left+right)//2
        #print(smi,a[mi],x)
        
        if a[mi]<=x:
            left = mi+1
            i+=1
            
        else:

            right = mi
            j+=1
        #print(left,right,"----")
    #print(i-1,j)
    if left>0 and a[left-1]==x:
        return i-1, j
    else:
        return -1, -1
def nCr(n, r):
 
    return (fact(n) / (fact(r)
                * fact(n - r)))
 
# Returns factorial of n
def fact(n):
 
    res = 1
     
    for i in range(2, n+1):
        res = res * i
         
    return res
n,k = lint()
dp=[[1]*(n+1) for _ in range(k+1)]

for i in range(1,k):
    
    for j in range(2,n+1):
        tmp=j
        while tmp <= n:
            dp[i+1][tmp] = (dp[i+1][tmp] + dp[i][j]) % (10**9+7)
            tmp += j
print((sum(dp[k])-1)%(10**9 +7))