#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(ll i=0;i<n;i++)
#define all(v) v.begin(), v.end()
typedef long long ll;
typedef pair<ll,ll> P;
typedef vector<ll> vec;
typedef vector<vec> mat;
ll mod=1e9+7,dp[110][200000];
int main(){
    ll n,k;
    cin>>n>>k;
    fill(dp[0],dp[0]+2*ll(sqrt(n))+1,0);
    dp[0][1]=1;
    ll count=0;
    for(int i=1;n/i>sqrt(n);i++) count++;
    count+=ll(sqrt(n));
    rep(i,k+1){
        ll a=1,sum=0,b=count;
        for(;n/a>sqrt(n);a++) (sum+=dp[i][a])%=mod,dp[i+1][b]=sum,b--;
        for(int j=sqrt(n);j>=1;j--){
            ll x=n/j-n/(j+1);
            (sum+=dp[i][a]*x)%=mod;
            dp[i+1][b]=sum,b--;
            a++;
        }
        
    }
    cout<<dp[k+1][1]<<"\n";
}
### Prompt

Please formulate 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;
#define rep(i,n) for(ll i=0;i<n;i++)
#define all(v) v.begin(), v.end()
typedef long long ll;
typedef pair<ll,ll> P;
typedef vector<ll> vec;
typedef vector<vec> mat;
ll mod=1e9+7,dp[110][200000];
int main(){
    ll n,k;
    cin>>n>>k;
    fill(dp[0],dp[0]+2*ll(sqrt(n))+1,0);
    dp[0][1]=1;
    ll count=0;
    for(int i=1;n/i>sqrt(n);i++) count++;
    count+=ll(sqrt(n));
    rep(i,k+1){
        ll a=1,sum=0,b=count;
        for(;n/a>sqrt(n);a++) (sum+=dp[i][a])%=mod,dp[i+1][b]=sum,b--;
        for(int j=sqrt(n);j>=1;j--){
            ll x=n/j-n/(j+1);
            (sum+=dp[i][a]*x)%=mod;
            dp[i+1][b]=sum,b--;
            a++;
        }
        
    }
    cout<<dp[k+1][1]<<"\n";
}
```