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(10^9 + 7).

The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).

Output a single integer — the number of good sequences of length k modulo 1000000007 (10^9 + 7).

In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3].

题意：

给出这么多数[1,n],问能形成长为l的数列满足b[i]|b[i+1]的方案数;

思路：

dp[i][j]表示长为i,以j结尾的方案数,

dp[i+1][j]=∑dp[i][k],k为j的因数;

AC代码：

#include <bits/stdc++.h>

using namespace std;

#define Riep(n) for(int i=1;i<=n;i++)

#define Riop(n) for(int i=0;i<n;i++)

#define Rjep(n) for(int j=1;j<=n;j++)

#define Rjop(n) for(int j=0;j<n;j++)

#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef long long LL;

const LL mod=1e9+;

const double PI=acos(-1.0);

const int inf=0x3f3f3f3f;

const int N=2e5+;

LL dp[][];

int main()

{

        int n,k;

        scanf("%d%d",&n,&k);

        for(int i=;i<=n;i++)

        {

            dp[][i]=;

        }

        for(int i=;i<=k;i++)

        {

            Rjep(n)

            {

                for(int x=;x*j<=;x++)

                {

                    dp[i+][x*j]+=dp[i][j];

                    dp[i+][x*j]%=mod;

                }

            }

        }

        LL ans=;

        for(int i=;i<=n;i++)

        {

            ans+=dp[k][i];

            ans%=mod;

        }

        cout<<ans<<"\n";

    return ;

}