#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <cmath>
#include <map>
#include <queue>
#include <iomanip>
#include <set>
#define MOD 1000000007
#define mkp make_pair
typedef long long ll;
using namespace std;

ll N,K;

ll dp[2][100010];

ll r[100010];
ll p[100010];

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

  int cnt=1;
  for(ll i=1;i<=N;){
      int x=N/i;
      int a=N/x;
      int b=N/(x+1)+1;
      if(i==1) b=1;

      r[cnt]=a-b+1;
      i+=(a-b+1);
      cnt++;
  }

  cnt--;
  for(int i=0;i<=cnt;i++) dp[0][i]=1;

  int t=0;
  for(int i=1;i<=K;i++){
      for(int j=1;j<=cnt;j++){
          dp[(t+1)%2][j]=dp[t][cnt-j+1]*r[j]%MOD;
      }

      t^=1;
      for(int j=2;j<=cnt;j++) dp[t][j]=(dp[t][j]+dp[t][j-1])%MOD;
  }

  ll ans=dp[t][cnt];

  cout<<ans<<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 <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <cmath>
#include <map>
#include <queue>
#include <iomanip>
#include <set>
#define MOD 1000000007
#define mkp make_pair
typedef long long ll;
using namespace std;

ll N,K;

ll dp[2][100010];

ll r[100010];
ll p[100010];

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

  int cnt=1;
  for(ll i=1;i<=N;){
      int x=N/i;
      int a=N/x;
      int b=N/(x+1)+1;
      if(i==1) b=1;

      r[cnt]=a-b+1;
      i+=(a-b+1);
      cnt++;
  }

  cnt--;
  for(int i=0;i<=cnt;i++) dp[0][i]=1;

  int t=0;
  for(int i=1;i<=K;i++){
      for(int j=1;j<=cnt;j++){
          dp[(t+1)%2][j]=dp[t][cnt-j+1]*r[j]%MOD;
      }

      t^=1;
      for(int j=2;j<=cnt;j++) dp[t][j]=(dp[t][j]+dp[t][j-1])%MOD;
  }

  ll ans=dp[t][cnt];

  cout<<ans<<endl;


  return 0;
}

```