#include <bits/stdc++.h>
using namespace std;
#define modulo 1000000007
#define mod(mod_x) ((((long long)mod_x)+modulo)%modulo)
#define Inf 1000000000

int main(){
	int N,K;
	cin>>N>>K;
	set<int> S;
	for(int i=1;i*i<=N;i++){
		int a = N/i;
		S.insert(a);
		a = N/a;
		S.insert(a);
	}
	vector<int> V;
	int last = 0;
	for(auto a:S){
		V.push_back(a-last);
		last=a;
	}
	
	vector<vector<int>> dp(K,vector<int>(V.size(),0));
	vector<vector<int>> sum(K,vector<int>(V.size(),0));
	
	for(int i=0;i<K;i++){
		for(int j=0;j<V.size();j++){
			dp[i][j] = V[j];
			if(i!=0){
				dp[i][j] = mod(dp[i][j]*sum[i-1][V.size()-1-j]);
			}
			
			
			sum[i][j] = dp[i][j];
			if(j!=0)sum[i][j] = mod(sum[i][j]+sum[i][j-1]);
		}
	}
	/*
	for(int i=0;i<K;i++){
		for(int j=0;j<V.size();j++){
			cout<<dp[i][j]<<',';
		}
		cout<<endl;
	}*/
	
	cout<<sum[K-1][V.size()-1]<<endl;
	
	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>
using namespace std;
#define modulo 1000000007
#define mod(mod_x) ((((long long)mod_x)+modulo)%modulo)
#define Inf 1000000000

int main(){
	int N,K;
	cin>>N>>K;
	set<int> S;
	for(int i=1;i*i<=N;i++){
		int a = N/i;
		S.insert(a);
		a = N/a;
		S.insert(a);
	}
	vector<int> V;
	int last = 0;
	for(auto a:S){
		V.push_back(a-last);
		last=a;
	}
	
	vector<vector<int>> dp(K,vector<int>(V.size(),0));
	vector<vector<int>> sum(K,vector<int>(V.size(),0));
	
	for(int i=0;i<K;i++){
		for(int j=0;j<V.size();j++){
			dp[i][j] = V[j];
			if(i!=0){
				dp[i][j] = mod(dp[i][j]*sum[i-1][V.size()-1-j]);
			}
			
			
			sum[i][j] = dp[i][j];
			if(j!=0)sum[i][j] = mod(sum[i][j]+sum[i][j-1]);
		}
	}
	/*
	for(int i=0;i<K;i++){
		for(int j=0;j<V.size();j++){
			cout<<dp[i][j]<<',';
		}
		cout<<endl;
	}*/
	
	cout<<sum[K-1][V.size()-1]<<endl;
	
	return 0;
}

```