#include<bits/stdc++.h>
#define ll long long
#define ld long double
#define fi first
#define se second
#define pb push_back
#define mod 1000000007
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define repf(i,a,b) for(ll i=a;i<b;i++)
#define repb(i,a,b) for(ll i=a;i>=b;i--)
#define all(v) v.begin(),v.end()

using namespace std;

int main()
{
    IOS;
    int n,k;
    cin>>n>>k;
    int m=sqrt(n)+1;
    ll a[k+1][m+1];
    ll b[k+1][m+1];
    repf(i,1,m+1) 
    {
        a[1][i]=min(i,(ll)n);
        b[1][i]=n/i;
    }
    repf(i,2,k+1)
    {
        a[i][0]=0;
        repf(j,1,m+1)
        {
            a[i][j]=(a[i][j-1]+b[i-1][j])%mod;
        }
        b[i][m]=a[i][n/m];
        repb(j,m-1,1)
        {
            b[i][j]=(b[i][j+1]+(a[i-1][j]*(n/j-n/(j+1)))%mod)%mod;
        }
    }
    cout<<b[k][1]<<endl;
}
 
### 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>
#define ll long long
#define ld long double
#define fi first
#define se second
#define pb push_back
#define mod 1000000007
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define repf(i,a,b) for(ll i=a;i<b;i++)
#define repb(i,a,b) for(ll i=a;i>=b;i--)
#define all(v) v.begin(),v.end()

using namespace std;

int main()
{
    IOS;
    int n,k;
    cin>>n>>k;
    int m=sqrt(n)+1;
    ll a[k+1][m+1];
    ll b[k+1][m+1];
    repf(i,1,m+1) 
    {
        a[1][i]=min(i,(ll)n);
        b[1][i]=n/i;
    }
    repf(i,2,k+1)
    {
        a[i][0]=0;
        repf(j,1,m+1)
        {
            a[i][j]=(a[i][j-1]+b[i-1][j])%mod;
        }
        b[i][m]=a[i][n/m];
        repb(j,m-1,1)
        {
            b[i][j]=(b[i][j+1]+(a[i-1][j]*(n/j-n/(j+1)))%mod)%mod;
        }
    }
    cout<<b[k][1]<<endl;
}
 
```