#include <algorithm>
#include <iostream>
#include <vector>

using namespace std;

typedef long long ll;

#define MOD 1000000007

ll dp[101][100000];
ll sum[101][100000];

int main() {
  int n, k;
  cin >> n >> k;
  vector<int> vec;
  vec.push_back(0);
  for (int i = 1; i * i <= n; ++i) {
    vec.push_back(i);
    if (i != n / i) {
      vec.push_back(n / i);
    }
  }
  sort(vec.begin(), vec.end());
  int m = vec.size() - 1;
  for (int i = 1; i <= m; ++i) {
    dp[1][i] = (vec[i] - vec[i - 1] + MOD) % MOD;
    sum[1][i] = (sum[1][i - 1] + dp[1][i]) % MOD;
  }
  for (int i = 2; i <= k; ++i) {
    for (int j = 1; j <= m; ++j) {
      dp[i][j] = sum[i - 1][m - (j - 1)];
      dp[i][j] = (dp[i][j] * (vec[j] - vec[j - 1] + MOD) % MOD) % MOD;
      sum[i][j] = (sum[i][j - 1] + dp[i][j]) % MOD;
    }
  }
  cout << sum[k][m] << endl;
  return 0;
}
### 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 <algorithm>
#include <iostream>
#include <vector>

using namespace std;

typedef long long ll;

#define MOD 1000000007

ll dp[101][100000];
ll sum[101][100000];

int main() {
  int n, k;
  cin >> n >> k;
  vector<int> vec;
  vec.push_back(0);
  for (int i = 1; i * i <= n; ++i) {
    vec.push_back(i);
    if (i != n / i) {
      vec.push_back(n / i);
    }
  }
  sort(vec.begin(), vec.end());
  int m = vec.size() - 1;
  for (int i = 1; i <= m; ++i) {
    dp[1][i] = (vec[i] - vec[i - 1] + MOD) % MOD;
    sum[1][i] = (sum[1][i - 1] + dp[1][i]) % MOD;
  }
  for (int i = 2; i <= k; ++i) {
    for (int j = 1; j <= m; ++j) {
      dp[i][j] = sum[i - 1][m - (j - 1)];
      dp[i][j] = (dp[i][j] * (vec[j] - vec[j - 1] + MOD) % MOD) % MOD;
      sum[i][j] = (sum[i][j - 1] + dp[i][j]) % MOD;
    }
  }
  cout << sum[k][m] << endl;
  return 0;
}
```