#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <climits>

#define MOD 1000000007

int cp[70505],lcp;
int f[70505],sum[70505];

int main()
{
	int N,M;scanf("%d%d",&M,&N);
	for(register int l=1,r;l<=M;l=r+1)
		r=M/(M/l),cp[++lcp]=r;
	for(register int j=1;j<=lcp;j++)
		f[j]=cp[j]-cp[j-1],
		sum[j]=(sum[j-1]+f[j])%MOD;
	for(register int i=2;i<=N;i++)
	{
		for(register int j=1;j<=lcp;j++)
			f[j]=(long long)sum[lcp-j+1]*(cp[j]-cp[j-1])%MOD;
		for(register int j=1;j<=lcp;j++)
			sum[j]=(sum[j-1]+f[j])%MOD;
	}
	printf("%d\n",sum[lcp]);
	return 0;
}
### 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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <climits>

#define MOD 1000000007

int cp[70505],lcp;
int f[70505],sum[70505];

int main()
{
	int N,M;scanf("%d%d",&M,&N);
	for(register int l=1,r;l<=M;l=r+1)
		r=M/(M/l),cp[++lcp]=r;
	for(register int j=1;j<=lcp;j++)
		f[j]=cp[j]-cp[j-1],
		sum[j]=(sum[j-1]+f[j])%MOD;
	for(register int i=2;i<=N;i++)
	{
		for(register int j=1;j<=lcp;j++)
			f[j]=(long long)sum[lcp-j+1]*(cp[j]-cp[j-1])%MOD;
		for(register int j=1;j<=lcp;j++)
			sum[j]=(sum[j-1]+f[j])%MOD;
	}
	printf("%d\n",sum[lcp]);
	return 0;
}
```