#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#define LL long long
const int mod=1e9+7;
using namespace std;
int n,k,f[110][33000][2],sum[110][33000][2],m;
int main()
{
	scanf("%d %d",&n,&k);
	m=sqrt(n);
	//printf("%d %d %d\n",m,n/m,n/(m+1));
	for(int i=0;i<=m;i++) sum[0][i][0]=1;
	for(int i=1;i<=k;i++)
	{
		for(int j=1;j<=m;j++) f[i][j][0]=(sum[i-1][m][0]+(sum[i-1][m][1]-sum[i-1][j-1][1])%mod)%mod;
		int j=m+1,r;
		while(j<=n)
		{
			r=n/(n/j);int t=n/j;
			//printf("%d %d\n",j,r);
			//if(n/j==n/(r+1)) printf("fuck\n");
			f[i][t][1]=(LL)(sum[i-1][t][0])*(r-j+1)%mod;
			j=r+1;
		}
		//printf("--------------------\n");
		for(int j=1;j<=m;j++)
		{
			sum[i][j][0]=(sum[i][j-1][0]+f[i][j][0])%mod;
			sum[i][j][1]=(sum[i][j-1][1]+f[i][j][1])%mod;
			//printf("%d %d\n",f[i][j][0],f[i][j][1]);
		}
	}
	int ans=(sum[k][m][0]+sum[k][m][1])%mod;
	printf("%d",(ans+mod)%mod);
}
### Prompt

Please create a solution in Cpp to 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<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#define LL long long
const int mod=1e9+7;
using namespace std;
int n,k,f[110][33000][2],sum[110][33000][2],m;
int main()
{
	scanf("%d %d",&n,&k);
	m=sqrt(n);
	//printf("%d %d %d\n",m,n/m,n/(m+1));
	for(int i=0;i<=m;i++) sum[0][i][0]=1;
	for(int i=1;i<=k;i++)
	{
		for(int j=1;j<=m;j++) f[i][j][0]=(sum[i-1][m][0]+(sum[i-1][m][1]-sum[i-1][j-1][1])%mod)%mod;
		int j=m+1,r;
		while(j<=n)
		{
			r=n/(n/j);int t=n/j;
			//printf("%d %d\n",j,r);
			//if(n/j==n/(r+1)) printf("fuck\n");
			f[i][t][1]=(LL)(sum[i-1][t][0])*(r-j+1)%mod;
			j=r+1;
		}
		//printf("--------------------\n");
		for(int j=1;j<=m;j++)
		{
			sum[i][j][0]=(sum[i][j-1][0]+f[i][j][0])%mod;
			sum[i][j][1]=(sum[i][j-1][1]+f[i][j][1])%mod;
			//printf("%d %d\n",f[i][j][0],f[i][j][1]);
		}
	}
	int ans=(sum[k][m][0]+sum[k][m][1])%mod;
	printf("%d",(ans+mod)%mod);
}
```