#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
int n,k;
ll dp[66666][111];
vector<int>v;
int main()
{
    scanf("%d%d",&n,&k);
    int l=1,r=1;
    while(r+1<=n)
    {
        v.push_back(r);
        l=r+1;r=n/(n/l);
    }
    v.push_back(r);
    sort(v.begin(),v.end());
    dp[0][1]=v[0];
    for(int i=1;i<v.size();i++)
        dp[i][1]=v[i]-v[i-1];
    ll ans=0;
    for(int i=2;i<=k;i++)
    {
        int r=v.size()-1;
        ll now=0;
        for(int j=0;j<v.size();j++)
        {
           now=(now+dp[j][i-1])%mod;
           while(v[r]>=n/v[j]&&r>=0)
           {
               dp[r][i]=now*(r==0?v[r]:v[r]-v[r-1])%mod;
               r--;
           }
        }
    }
    for(int i=0;i<v.size();i++)
        ans+=dp[i][k];
    printf("%lld\n",ans%mod);
}

### Prompt

Your task is to create 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;
typedef long long ll;
const int mod=1e9+7;
int n,k;
ll dp[66666][111];
vector<int>v;
int main()
{
    scanf("%d%d",&n,&k);
    int l=1,r=1;
    while(r+1<=n)
    {
        v.push_back(r);
        l=r+1;r=n/(n/l);
    }
    v.push_back(r);
    sort(v.begin(),v.end());
    dp[0][1]=v[0];
    for(int i=1;i<v.size();i++)
        dp[i][1]=v[i]-v[i-1];
    ll ans=0;
    for(int i=2;i<=k;i++)
    {
        int r=v.size()-1;
        ll now=0;
        for(int j=0;j<v.size();j++)
        {
           now=(now+dp[j][i-1])%mod;
           while(v[r]>=n/v[j]&&r>=0)
           {
               dp[r][i]=now*(r==0?v[r]:v[r]-v[r-1])%mod;
               r--;
           }
        }
    }
    for(int i=0;i<v.size();i++)
        ans+=dp[i][k];
    printf("%lld\n",ans%mod);
}

```