Codeforces 414b Mashmokh and ACM

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  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].

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;#define MOD1000000007#define N 2005int dp[N][N],n;void init(){int i,j,z,k;for(i=1;i<=N;i++){dp[1][i]=1;}for(i=1;i<=N;i++)     for(j=1;j<=N;j++)  for(z=j;z<=N;z+=j)   dp[i][z]=(dp[i][z]+dp[i-1][j])%MOD;   }int main(){int z,k,i,j;init();while(cin>>n>>k){   int s=0;    for(i=1;i<=n;i++)    {    s+=dp[k][i];    s%=MOD;}cout<<s<<endl;} return 0;}