#include <bits/stdc++.h>
#define pb push_back
#define int long long
#define mp make_pair
#define inf 1000000007
#define ll long long
using namespace std;
ll sma[101][32000],lar[101][32000];
signed main(){
	int n,k;
	cin>>n>>k;
	int s=0;
	for(int i=1;i<=32000;i++){
		if( i*i <= n)s=i;
	}
	sma[0][1] = 1;
	for(int i=2;i<=s;i++){
		sma[0][i] = sma[0][i-1];
	}
	for(int i=1;i<=k;i++){
		for(int j=1;j<=s;j++){
			sma[i][j] = sma[i-1][s] + lar[i-1][j];
			sma[i][j] %= inf;
		}
		for(int j=s;j>=1;j--){
			lar[i][j] = sma[i-1][j]*(min(n/j-n/(j+1), n/j-s));
			lar[i][j] %= inf;
		}
		for(int j=1;j<=s;j++){
			sma[i][j] += sma[i][j-1];
			sma[i][j] %= inf;
		}
		for(int j=s;j>=1;j--){
			lar[i][j] += lar[i][j+1];
			lar[i][j] %= inf;
		}
	}
	cout<<(sma[k][s]+lar[k][1])%inf<<endl;
	return 0;
}
### Prompt

Your task is to create a cpp solution 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 <bits/stdc++.h>
#define pb push_back
#define int long long
#define mp make_pair
#define inf 1000000007
#define ll long long
using namespace std;
ll sma[101][32000],lar[101][32000];
signed main(){
	int n,k;
	cin>>n>>k;
	int s=0;
	for(int i=1;i<=32000;i++){
		if( i*i <= n)s=i;
	}
	sma[0][1] = 1;
	for(int i=2;i<=s;i++){
		sma[0][i] = sma[0][i-1];
	}
	for(int i=1;i<=k;i++){
		for(int j=1;j<=s;j++){
			sma[i][j] = sma[i-1][s] + lar[i-1][j];
			sma[i][j] %= inf;
		}
		for(int j=s;j>=1;j--){
			lar[i][j] = sma[i-1][j]*(min(n/j-n/(j+1), n/j-s));
			lar[i][j] %= inf;
		}
		for(int j=1;j<=s;j++){
			sma[i][j] += sma[i][j-1];
			sma[i][j] %= inf;
		}
		for(int j=s;j>=1;j--){
			lar[i][j] += lar[i][j+1];
			lar[i][j] %= inf;
		}
	}
	cout<<(sma[k][s]+lar[k][1])%inf<<endl;
	return 0;
}
```