Codeforces Round #240 (Div. 1) B. Mashmokh and ACM DP

                                             B. Mashmokh and ACM

                                                                                            time limit per test

                                                                                            1 second

                                                                                            memory limit per test

                                                                                            256 megabytes

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

Note

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

题意：给你一个N，K,  表示从1到n，选取长度为k的序列A满足  Ai整除Ai-1

题解：dp[i][j]表示长度为j是最大为i的序列方案数

//1085422276
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<cmath>
#include<map>
#include<bitset>
#include<set>
#include<vector>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,127,sizeof(a));
#define memfy(a) memset(a,-1,sizeof(a))
#define TS printf("111111
");
#define FOR(i,a,b) for( int i=a;i<=b;i++)
#define FORJ(i,a,b) for(int i=a;i>=b;i--)
#define READ(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define mod 1000000007
#define maxn 2005
inline ll read()
{
    ll x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
//****************************************
ll  dp[maxn][maxn];

int main()
{

    int n=read(),m=read();
    mem(dp);
    FOR(i,1,n)dp[i][1]=1;
    FOR(k,1,m-1)
    FOR(i,1,n)
    { if(dp[i][k])
         for(int j=i;j<=n;j+=i)
         {
             dp[j][k+1]=(dp[j][k+1]+dp[i][k])%mod;
         }
    }
    ll ans=0;
    FOR(i,1,n)ans=(ans+dp[i][m])%mod;
    cout<<ans<<endl;
    return 0;
}