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

long long n=1e4,k,cont,mod=1e9+7ll,dp[100][65000];

vector<long long> v;

int main(){

    cin>>n>>k;
    
    v.push_back(0);
	for(long long i=1;i<=n;i++){
	    i=n/(n/i);
	    v.push_back(i);
    }
    
    for(int i=k-1;i>=0;i--){
        for(int j=v.size()-1;j>0;j--){
            
            if(i==k-1)dp[i][j]=v[j];
            else
            dp[i][j]=dp[i+1][v.size()-j];
            
            //cout<<i<<" "<<j<<" "<<v[j]<<endl;
            //cout<<dp[i][j]<<endl;
            dp[i][j]*=(v[v.size()-j]-v[v.size()-j-1])%mod;
            //cout<<dp[i][j]<<endl;
            dp[i][j]%=mod;
            dp[i][j]+=dp[i][j+1];
            dp[i][j]%=mod;
        }
    }
    
    
    cout<<dp[0][v.size()-1]<<"\n";

return 0;
}

### Prompt

Please create a solution in CPP 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 n=1e4,k,cont,mod=1e9+7ll,dp[100][65000];

vector<long long> v;

int main(){

    cin>>n>>k;
    
    v.push_back(0);
	for(long long i=1;i<=n;i++){
	    i=n/(n/i);
	    v.push_back(i);
    }
    
    for(int i=k-1;i>=0;i--){
        for(int j=v.size()-1;j>0;j--){
            
            if(i==k-1)dp[i][j]=v[j];
            else
            dp[i][j]=dp[i+1][v.size()-j];
            
            //cout<<i<<" "<<j<<" "<<v[j]<<endl;
            //cout<<dp[i][j]<<endl;
            dp[i][j]*=(v[v.size()-j]-v[v.size()-j-1])%mod;
            //cout<<dp[i][j]<<endl;
            dp[i][j]%=mod;
            dp[i][j]+=dp[i][j+1];
            dp[i][j]%=mod;
        }
    }
    
    
    cout<<dp[0][v.size()-1]<<"\n";

return 0;
}

```