#include <iostream>
#include <cmath>
using namespace std;

typedef long long ll;
ll mod=1e9+7;

ll modpow(ll a,ll b){
  ll ans=1;
  a%=mod;
  while(b){
    if(b&1) ans=ans*a%mod;
    a=a*a%mod;
    b>>=1;
  }
  return ans;
}

ll dp[100000][101];
ll sum[100000][101];

int main(){
  ll n,k;
  cin >> n >> k;
  ll t=sqrt(n);
  for(ll i=1;i<=t;i++) dp[i][1]=1;
  ll now=t+1;
  ll s;
  if((t+1)*t>n) s=t-1;
  else s=t;
  for(ll i=s;i>=1;i--){
    dp[now][1]=n/i-n/(i+1);
    now++;
  }
  for(ll i=2;i<=k;i++){
    for(ll j=1;j<now;j++){
      sum[j][i-1]=(sum[j-1][i-1]+dp[j][i-1])%mod;
    }
    for(ll j=1;j<now;j++){
      dp[j][i]=(sum[now-j][i-1]*dp[j][1])%mod;
    }
  }
  ll ans=0;
  for(ll i=1;i<now;i++){
    ans=(ans+dp[i][k])%mod;
  }
  cout << ans << endl;
}
### 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 <cmath>
using namespace std;

typedef long long ll;
ll mod=1e9+7;

ll modpow(ll a,ll b){
  ll ans=1;
  a%=mod;
  while(b){
    if(b&1) ans=ans*a%mod;
    a=a*a%mod;
    b>>=1;
  }
  return ans;
}

ll dp[100000][101];
ll sum[100000][101];

int main(){
  ll n,k;
  cin >> n >> k;
  ll t=sqrt(n);
  for(ll i=1;i<=t;i++) dp[i][1]=1;
  ll now=t+1;
  ll s;
  if((t+1)*t>n) s=t-1;
  else s=t;
  for(ll i=s;i>=1;i--){
    dp[now][1]=n/i-n/(i+1);
    now++;
  }
  for(ll i=2;i<=k;i++){
    for(ll j=1;j<now;j++){
      sum[j][i-1]=(sum[j-1][i-1]+dp[j][i-1])%mod;
    }
    for(ll j=1;j<now;j++){
      dp[j][i]=(sum[now-j][i-1]*dp[j][1])%mod;
    }
  }
  ll ans=0;
  for(ll i=1;i<now;i++){
    ans=(ans+dp[i][k])%mod;
  }
  cout << ans << endl;
}
```