#include <bits/stdc++.h>
using namespace std;

long long MOD=1000000007LL;

long long dp[101][64000];

int main(){
	int N,K;
	cin >> N >> K;

	int rN=sqrt(N);
	vector<long long> init;
	int c=0;
	for(int i=0;i<=rN;i++){
		if(i!=rN || i*(i+1)<=N)init.push_back(N-N/(i+1)-c);
		c+=N-N/(i+1)-c;
	}
	for(int i=1;i<=rN;i++)init.push_back(1);
	reverse(init.begin(),init.end());
	
	int M=init.size();
	for(int i=0;i<M;i++)dp[0][i+1]=init[i];
	
	for(int i=0;i<K-1;i++){
		vector<long long> sum(M+1,0);
		for(int j=0;j<M;j++)sum[j+1]=sum[j]+dp[i][j];
		for(auto &k:sum)k%=MOD;
		for(int j=1;j<M;j++)dp[i+1][j]=sum[M-j+1]*init[j-1];
	}
	
	long long ans=0;
	for(int j=1;j<M;j++)ans+=dp[K-1][j];
	cout << ans%MOD << endl;
	
	/*for(int i=0;i<K;i++){
		for(int j=0;j<M;j++)cout << dp[i][j] << " ";
		cout << endl;
	}*/
	return 0;
}
### Prompt

Your challenge is to write 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>
using namespace std;

long long MOD=1000000007LL;

long long dp[101][64000];

int main(){
	int N,K;
	cin >> N >> K;

	int rN=sqrt(N);
	vector<long long> init;
	int c=0;
	for(int i=0;i<=rN;i++){
		if(i!=rN || i*(i+1)<=N)init.push_back(N-N/(i+1)-c);
		c+=N-N/(i+1)-c;
	}
	for(int i=1;i<=rN;i++)init.push_back(1);
	reverse(init.begin(),init.end());
	
	int M=init.size();
	for(int i=0;i<M;i++)dp[0][i+1]=init[i];
	
	for(int i=0;i<K-1;i++){
		vector<long long> sum(M+1,0);
		for(int j=0;j<M;j++)sum[j+1]=sum[j]+dp[i][j];
		for(auto &k:sum)k%=MOD;
		for(int j=1;j<M;j++)dp[i+1][j]=sum[M-j+1]*init[j-1];
	}
	
	long long ans=0;
	for(int j=1;j<M;j++)ans+=dp[K-1][j];
	cout << ans%MOD << endl;
	
	/*for(int i=0;i<K;i++){
		for(int j=0;j<M;j++)cout << dp[i][j] << " ";
		cout << endl;
	}*/
	return 0;
}
```