#include<iostream>
#include<stdio.h>
#include<queue>
#include<vector>
#define ll long long
using namespace std;
const int inf=31630;
const int maxn=101;
const int mod=1e9+7;
int dp[maxn][2*inf];
struct node {
	int x,y;
};
vector<node>a;
void add(int x,int y){
	x+=y;
	while(x>=mod)x-=mod;
}
int main() {
	int n,k;
	scanf("%d %d",&n,&k);
	int x=1,y;
	a.push_back((node)<%0,0%>);
	while(x<=n){
		int t=n/x;
		y=n/t;
		a.push_back((node)<%x,y%>);
		x=y+1;
//		printf("x : %d,y : %d\n",x,y);
	}
	int tot=a.size()-1;
	for(int i=1; i<=tot; i++) {
		dp[1][i]=a[i].y-a[i].x+1;
		dp[1][i]+=dp[1][i-1];
	}
	for(int i=2;i<=k;i++)
		for(int j=1;j<=tot;j++)
			dp[i][j]=(1LL*dp[i-1][tot-j+1]*(a[j].y-a[j].x+1)%mod+dp[i][j-1]%mod)%mod;
	ll ans=1LL*dp[k][tot];
	printf("%lld\n",ans);
	return 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

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<iostream>
#include<stdio.h>
#include<queue>
#include<vector>
#define ll long long
using namespace std;
const int inf=31630;
const int maxn=101;
const int mod=1e9+7;
int dp[maxn][2*inf];
struct node {
	int x,y;
};
vector<node>a;
void add(int x,int y){
	x+=y;
	while(x>=mod)x-=mod;
}
int main() {
	int n,k;
	scanf("%d %d",&n,&k);
	int x=1,y;
	a.push_back((node)<%0,0%>);
	while(x<=n){
		int t=n/x;
		y=n/t;
		a.push_back((node)<%x,y%>);
		x=y+1;
//		printf("x : %d,y : %d\n",x,y);
	}
	int tot=a.size()-1;
	for(int i=1; i<=tot; i++) {
		dp[1][i]=a[i].y-a[i].x+1;
		dp[1][i]+=dp[1][i-1];
	}
	for(int i=2;i<=k;i++)
		for(int j=1;j<=tot;j++)
			dp[i][j]=(1LL*dp[i-1][tot-j+1]*(a[j].y-a[j].x+1)%mod+dp[i][j-1]%mod)%mod;
	ll ans=1LL*dp[k][tot];
	printf("%lld\n",ans);
	return 0;
}
```