#include<bits/stdc++.h>
#define mxn 1000010
#define LL long long
#define pb push_back
#define ppb pop_back
#define pf push_front
#define pii pair<int,int> 
#define mp make_pair
#define fr first
#define sc second
#define pb push_back
#define node(a,b,c) (node){a,b,c}
using namespace std;
const int inf=1e9,mod=1e9+7;
int n,m,sl,fh,top,ans,f[mxn],dp[110][mxn];
int rd()
{
	sl=0;fh=1;
	char ch=getchar();
	while(ch<'0'||'9'<ch) {if(ch=='-') fh=-1; ch=getchar();}
	while('0'<=ch&&ch<='9') sl=sl*10+ch-'0',ch=getchar();
	return sl*fh;
}
void upd(int &x,int y) {x+=y; if(x>=mod) x-=mod;}
int main()
{
	n=rd();m=rd();
	for(int r,l=1;l<=n;l=r+1)
	{
		r=min(n,n/(n/l));
		f[++top]=r-l+1;
	}
	for(int i=1;i<=top;++i) dp[1][i]=f[i];
	for(int i=2;i<=m;++i)
		for(int j=1;j<=top;++j)
			upd(dp[i-1][j],dp[i-1][j-1]),dp[i][top-j+1]=1ll*dp[i-1][j]*f[top-j+1]%mod;
	for(int i=1;i<=top;++i) upd(ans,dp[m][i]);
	printf("%d\n",ans);
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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>
#define mxn 1000010
#define LL long long
#define pb push_back
#define ppb pop_back
#define pf push_front
#define pii pair<int,int> 
#define mp make_pair
#define fr first
#define sc second
#define pb push_back
#define node(a,b,c) (node){a,b,c}
using namespace std;
const int inf=1e9,mod=1e9+7;
int n,m,sl,fh,top,ans,f[mxn],dp[110][mxn];
int rd()
{
	sl=0;fh=1;
	char ch=getchar();
	while(ch<'0'||'9'<ch) {if(ch=='-') fh=-1; ch=getchar();}
	while('0'<=ch&&ch<='9') sl=sl*10+ch-'0',ch=getchar();
	return sl*fh;
}
void upd(int &x,int y) {x+=y; if(x>=mod) x-=mod;}
int main()
{
	n=rd();m=rd();
	for(int r,l=1;l<=n;l=r+1)
	{
		r=min(n,n/(n/l));
		f[++top]=r-l+1;
	}
	for(int i=1;i<=top;++i) dp[1][i]=f[i];
	for(int i=2;i<=m;++i)
		for(int j=1;j<=top;++j)
			upd(dp[i-1][j],dp[i-1][j-1]),dp[i][top-j+1]=1ll*dp[i-1][j]*f[top-j+1]%mod;
	for(int i=1;i<=top;++i) upd(ans,dp[m][i]);
	printf("%d\n",ans);
	return 0;
}
```