#include <bits/stdc++.h>

using namespace std;
using i64 = int64_t;

const i64 MOD = 1000000007;

int main(){
  i64 N, K;
  cin >> N >> K;
  i64 NN = 1;
  while(NN * NN <= N)NN++;
  NN--;
  deque<i64> D;
  D.push_back(1);
  if((N / NN) != NN)D.push_front(N / NN - N / (NN + 1));
  while(NN >= 2){
    NN--;
    D.push_back(1);
    D.push_front(N / NN - N / (NN + 1));
  }
  deque<i64> C = D;
  K--;
//  for(auto&& x:D){
//	cerr << x << " ";
//  }cerr << endl;
  while(K--){
    reverse(D.begin(), D.end());
    D[0] %= MOD;
    for(int i=0;i<(int)D.size()-1;i++){
      D[i+1] += D[i];
      D[i+1] %= MOD;
    }
    for(int i=0;i<(int)D.size();i++){
      D[i] *= C[i];
      D[i] %= MOD;
    }
  }
  i64 ans = 0;
  for(auto&& x:D){
    ans += x;
    ans %= MOD;
  }
    cout << ans << endl;
}

### 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 i64 = int64_t;

const i64 MOD = 1000000007;

int main(){
  i64 N, K;
  cin >> N >> K;
  i64 NN = 1;
  while(NN * NN <= N)NN++;
  NN--;
  deque<i64> D;
  D.push_back(1);
  if((N / NN) != NN)D.push_front(N / NN - N / (NN + 1));
  while(NN >= 2){
    NN--;
    D.push_back(1);
    D.push_front(N / NN - N / (NN + 1));
  }
  deque<i64> C = D;
  K--;
//  for(auto&& x:D){
//	cerr << x << " ";
//  }cerr << endl;
  while(K--){
    reverse(D.begin(), D.end());
    D[0] %= MOD;
    for(int i=0;i<(int)D.size()-1;i++){
      D[i+1] += D[i];
      D[i+1] %= MOD;
    }
    for(int i=0;i<(int)D.size();i++){
      D[i] *= C[i];
      D[i] %= MOD;
    }
  }
  i64 ans = 0;
  for(auto&& x:D){
    ans += x;
    ans %= MOD;
  }
    cout << ans << endl;
}

```