#include<bits/stdc++.h>
using namespace std;
using ll = long long;
const ll mof = 1e9+7;
int solve(int n,int k)
{
    vector<int> facts;
    int sqrtn = static_cast<int>(sqrt(n));
    for(int i=1;i<=sqrtn;++i)
    {
        facts.push_back(i);
        if(n/i != i)
            facts.push_back(n/i);
    }
    facts.push_back(0);
    sort(facts.begin(),facts.end());
    vector<ll> dp0,dp1;
    dp1.resize(facts.size());
    for(int i=1;i<facts.size();++i)
        dp1[i] = (dp1[i-1] + facts[i] - facts[i-1]) % mof;
    for(int i=2;i<=k;++i)
    {
        dp0 = dp1;
        dp1.resize(dp0.size(),0);
        for(int j=1;j<dp0.size();++j)
        {
            dp1[j] = dp0[facts.size() - j] * (facts[j] - facts[j-1]) % mof;
            dp1[j] += dp1[j-1];
            dp1[j] %= mof;
        }
    }
    return dp1.back();
}
int main()
{
    int n,k;
    cin>>n>>k;
    cout<<solve(n,k)<<endl;
}
### Prompt

In cpp, your task is to solve 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;
using ll = long long;
const ll mof = 1e9+7;
int solve(int n,int k)
{
    vector<int> facts;
    int sqrtn = static_cast<int>(sqrt(n));
    for(int i=1;i<=sqrtn;++i)
    {
        facts.push_back(i);
        if(n/i != i)
            facts.push_back(n/i);
    }
    facts.push_back(0);
    sort(facts.begin(),facts.end());
    vector<ll> dp0,dp1;
    dp1.resize(facts.size());
    for(int i=1;i<facts.size();++i)
        dp1[i] = (dp1[i-1] + facts[i] - facts[i-1]) % mof;
    for(int i=2;i<=k;++i)
    {
        dp0 = dp1;
        dp1.resize(dp0.size(),0);
        for(int j=1;j<dp0.size();++j)
        {
            dp1[j] = dp0[facts.size() - j] * (facts[j] - facts[j-1]) % mof;
            dp1[j] += dp1[j-1];
            dp1[j] %= mof;
        }
    }
    return dp1.back();
}
int main()
{
    int n,k;
    cin>>n>>k;
    cout<<solve(n,k)<<endl;
}
```