CodeForces 414B--Mashmokh and ACM (dp)

B. Mashmokh and ACM

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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).

Sample test(s)

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].

/*
dp[i][j]=dp[i的所有因子][j-1]的和；
dp[i][j]表示最高位为i，长度为j时的个数
*/

#include
#include
#include
#include
#include
using namespace std;
const int Mod=1e9+7;
const int M=2001+10;
long long dp[M][M];
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n,k,i,j;
    while(cin>>n>>k){
        memset(dp,0,sizeof(dp));
        dp[1][0]=1;
        for(i=1;i<=n;i++){
            for(j=1;j<=k;j++){
                for(int ii=1; ii*ii <= i;ii++){//求数i的所有因子，根号n的复杂度，
                    if(i%ii == 0 ){            //如果数ii是i的因子，那么i/ii也是i的因子。不优化会超时
                        dp[i][j] = (dp[i][j] + dp[ii][j-1])%Mod;
                        if(ii != i/ii)
                            dp[i][j] = (dp[i][j] + dp[i/ii][j-1])%Mod;
                    }
                }
            }
        }
        long long res=0;
        for(i=1; i<=n; i++)
            res+=dp[i][k]%Mod;
        cout<