#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
using T = tuple<ll,int,int>;
const ll inf = numeric_limits<ll>::max()/3;
const ll mod = 1e9 + 7;

int main(void){
  ll n,k;
  cin >> n >> k;
  set<ll> bound;
  bound.insert(0);
  for(ll i = 1;i * i <= n;++i){
    bound.insert(i);
    bound.insert(n/i);
  }
  auto elem = vector< ll >(bound.begin(),bound.end());
  int sz = bound.size() - 1;
  auto calc = vector< ll >(sz);
  for(int i = 0;i < sz;++i){
    calc[i] = (elem[i+1] - elem[i]) % mod;
  }
  for(int i = 0;i < k-1;++i){
    auto next = vector< ll >(sz,0);
    ll acc = 0LL;
    for(int j = sz-1;j >= 0;--j){
      acc += calc[sz-1-j];
      next[j] = acc * (elem[j+1] - elem[j]);
      next[j] %= mod;
    }
    calc = next;
  }
  ll res = 0LL;
  for(int i = 0;i < sz;++i){
    res += calc[i];
    res %= mod;
  }
  cout << res << endl;
  return 0;
}

### Prompt

Generate 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;
using ll = long long;
using P = pair<ll, ll>;
using T = tuple<ll,int,int>;
const ll inf = numeric_limits<ll>::max()/3;
const ll mod = 1e9 + 7;

int main(void){
  ll n,k;
  cin >> n >> k;
  set<ll> bound;
  bound.insert(0);
  for(ll i = 1;i * i <= n;++i){
    bound.insert(i);
    bound.insert(n/i);
  }
  auto elem = vector< ll >(bound.begin(),bound.end());
  int sz = bound.size() - 1;
  auto calc = vector< ll >(sz);
  for(int i = 0;i < sz;++i){
    calc[i] = (elem[i+1] - elem[i]) % mod;
  }
  for(int i = 0;i < k-1;++i){
    auto next = vector< ll >(sz,0);
    ll acc = 0LL;
    for(int j = sz-1;j >= 0;--j){
      acc += calc[sz-1-j];
      next[j] = acc * (elem[j+1] - elem[j]);
      next[j] %= mod;
    }
    calc = next;
  }
  ll res = 0LL;
  for(int i = 0;i < sz;++i){
    res += calc[i];
    res %= mod;
  }
  cout << res << endl;
  return 0;
}

```