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 JAVA):

Solution:
import java.util.*;
import java.io.*;
public class Mashmokh2{
public static long mod = 1000000007L;
public static long bn[][];
public static int prim[] = new int[2001];
public static void main(String args[]) throws IOException{
BufferedReader lector = new BufferedReader(new InputStreamReader(System.in));
String tmp[] = lector.readLine().split(" ");
for(int n = 2;n<=Math.sqrt(2001);n++)
if(prim[n]==0)
for(int k = n*n;k<2001;k+=n)prim[k]=n;
int a  = Integer.parseInt(tmp[0]);
int b= Integer.parseInt(tmp[1]);
bn = new long[b+11][b+11];
bn[0][0]=1L;
for(int n = 1;n<bn.length;n++){
bn[n][0]=1L;
bn[n][n]=1L;
for(int m = 1;m<n;m++)bn[n][m]=(bn[n-1][m]+bn[n-1][m-1])%mod;
}
long res = 0;
for(int n = 1;n<=a;n++)
res = (res+hagale(n,b))%mod;
System.out.println(res);
}
public static long hagale(int a,int b){
long res = 1L;
if(a==1)return 1L;
do{
int q = prim[a];
//System.out.println(q);
if(q==0){
res = (res*b)%mod;
break;
}
int ex = 0;
while(a%q==0 && ++ex>0)a/=q;
res = (res*bn[ex+b-1][b-1])%mod;
}while(a!=1);
return res;
}
}