#include<bits/stdc++.h>
using namespace std;
const int N=110,T=72010,p=1e9+7;
int add(int x,int y){return(x+y)%p;}
int mul(int x,int y){return 1ll*x*y%p;}
int n,k;
int f[N][T],siz[T];
vector<int>s;
#define mp(x) (lower_bound(s.begin(),s.end(),x)-s.begin()+1)
int main(){
//	freopen("1.in","r",stdin);
    cin>>n>>k;
    for(int i=1,l=1,r;l<=n;l=r+1,++i){
        r=n/(n/l);
        f[1][i]=r-l+1;
        siz[i]=r-l+1;
	    s.push_back(n/l);
	}
	sort(s.begin(),s.end());
	int len=s.size();
	for(int i=2;i<=k;++i){
		for(int j=1;j<=len;++j)f[i-1][j]=add(f[i-1][j],f[i-1][j-1]);
	    for(int j=1;j<=len;++j){
	        f[i][j]=mul(f[i-1][mp(n/s[j-1])],siz[j]);
		}
	}
	int ans=0;
	for(int i=1;i<=len;++i)ans=add(ans,f[k][i]);
	cout<<ans<<endl;
//	cout<<s.size()<<endl;
//	for(vector<int>::iterator it=s.begin();it!=s.end();++it)cout<<(*it)<<' ';puts("");
}
### Prompt

Create a solution in CPP for the following problem:

Find the number of sequences of length K consisting of positive integers such that the product of any two adjacent elements is at most N, modulo 10^9+7.

Constraints

* 1\leq N\leq 10^9
* ~~1~~ 2\leq K\leq 100 (fixed at 21:33 JST)
* N and K are integers.

Input

Input is given from Standard Input in the following format:


N K


Output

Print the number of sequences, modulo 10^9+7.

Examples

Input

3 2


Output

5


Input

10 3


Output

147


Input

314159265 35


Output

457397712

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
const int N=110,T=72010,p=1e9+7;
int add(int x,int y){return(x+y)%p;}
int mul(int x,int y){return 1ll*x*y%p;}
int n,k;
int f[N][T],siz[T];
vector<int>s;
#define mp(x) (lower_bound(s.begin(),s.end(),x)-s.begin()+1)
int main(){
//	freopen("1.in","r",stdin);
    cin>>n>>k;
    for(int i=1,l=1,r;l<=n;l=r+1,++i){
        r=n/(n/l);
        f[1][i]=r-l+1;
        siz[i]=r-l+1;
	    s.push_back(n/l);
	}
	sort(s.begin(),s.end());
	int len=s.size();
	for(int i=2;i<=k;++i){
		for(int j=1;j<=len;++j)f[i-1][j]=add(f[i-1][j],f[i-1][j-1]);
	    for(int j=1;j<=len;++j){
	        f[i][j]=mul(f[i-1][mp(n/s[j-1])],siz[j]);
		}
	}
	int ans=0;
	for(int i=1;i<=len;++i)ans=add(ans,f[k][i]);
	cout<<ans<<endl;
//	cout<<s.size()<<endl;
//	for(vector<int>::iterator it=s.begin();it!=s.end();++it)cout<<(*it)<<' ';puts("");
}
```