Codeforces 414B Mashmokh and ACM (DP)

Codeforces 414B Mashmokh and ACM (DP)

Mashmokh's boss, Bimokh, didn't like Mashmokh. so he fired him. mashmokh decided to go to university and participant ipate 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 sequenceLIntegersB1 ,?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 allI(1? ≤?I? ≤?L? -? 1 ).

GivenNAndKFind the number of good sequences of lengthK. As the answer can be rather large print it modulo 1000000007 (109? +? 7 ).

Input

The first line of input contains two space-separated integersN,?K(1? ≤?N,?K? ≤? 2000 ).

Output

Output a single integer-the number of good sequences of lengthKModulo 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].

The question is the DP about the counting problem. If dp [I] [j] is set, it indicates the number of solutions whose end length is j with I: dp [I] [j] = sum (dp [k] [J-1]) (I % k = 0) Special dp [I] [1] = 1;

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           using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )typedef long long LL;typedef pair
           
            pil;const int INF = 0x3f3f3f3f;const int MOD=1e9+7;const int maxn=2000+100;LL dp[maxn][maxn];int n,k;int main(){ while(cin>>n>>k) { CLEAR(dp,0); REPF(i,1,n) dp[i][1]=1; for(int j=1;j<=n;j++) for(int kk=j;kk<=n;kk+=j) for(int i=2;i<=k;i++) dp[kk][i]=(dp[kk][i]+dp[j][i-1])%MOD; LL ans=0; REPF(i,1,n) ans=(ans+dp[i][k])%MOD;