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
import math as mt
input=sys.stdin.buffer.readline  
t=1
mod=10**9+7

def getd() : 
    l=[]  
    # Note that this loop runs till square root 
    l.append([0])
    l.append([1])
    for n in range(2,20001):
        i = 1
        l1=[]
        while i <= mt.sqrt(n): 
              
            if (n % i == 0) : 
                  
                # If divisors are equal, print only one 
                if (n // i == i) : 
                    l1.append(i)
                     
                else : 
                    l1.append(i)
                    l1.append(n//i)
            i = i + 1
        l.append(l1)        
    return l    
l=getd()   
#print(l[:7])
for __ in range(t):
    n,k=map(int,input().split())
    dp=[[0 for j in range(n+1)] for i in range(k+1)]
    for j in range(1,n+1):
        dp[1][j]=1
    for i in range(1,k+1):
        dp[i][1]=1
    for i in range(2,k+1):
        for j in range(2,n+1):
            for j1 in l[j]:
                dp[i][j]=(dp[i][j]%mod+dp[i-1][j1]%mod)%mod
    suma=0
    for j in range(1,n+1):
        suma=(suma+dp[k][j])%mod
        #suma%mod
    print(suma%mod)