#include<bits/stdc++.h>
using namespace std;
const int N = 66666, MOD = 1e9 + 7; // N > 2 * sqrt(n) 就可
void add(int &x, int y) {
  x += y;
  if(x >= MOD) x -= MOD;
}
vector<int> v;
int dp[2][N], len[N]; // len[] 记录每一个一起计算的部分的长度
int main() {
  ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  int n, k; cin >> n >> k;
  for(int i = 1, j; i <= n; i = j + 1) {
    j = n / (n / i); // 边界值
    v.push_back(i);
  }
  v.push_back(n + 1);
  for(int i = 0; i + 1 < int(v.size()); i++) {
    len[i] = v[i + 1] - v[i];
  }
  const int m = v.size();
  for(int i = 1; i < m; i++) dp[0][i] = 1;
  for(int q = 1; q <= k; q++) {
    int cur = 1, whe = m - 1;
    for(int i = 1, j; i <= n; i = j + 1, cur++, whe--) {
      j = n / (n / i); // 边界值
      dp[q & 1][cur] = 1ll * dp[~q & 1][whe] * len[cur - 1] % MOD; // dp转移
    }
    for(int i = 1; i < m; i++) add(dp[q & 1][i], dp[q & 1][i - 1]); // 求前缀和
  }
  cout << dp[k & 1][m - 1] << "\n";
}
### 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;
const int N = 66666, MOD = 1e9 + 7; // N > 2 * sqrt(n) 就可
void add(int &x, int y) {
  x += y;
  if(x >= MOD) x -= MOD;
}
vector<int> v;
int dp[2][N], len[N]; // len[] 记录每一个一起计算的部分的长度
int main() {
  ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  int n, k; cin >> n >> k;
  for(int i = 1, j; i <= n; i = j + 1) {
    j = n / (n / i); // 边界值
    v.push_back(i);
  }
  v.push_back(n + 1);
  for(int i = 0; i + 1 < int(v.size()); i++) {
    len[i] = v[i + 1] - v[i];
  }
  const int m = v.size();
  for(int i = 1; i < m; i++) dp[0][i] = 1;
  for(int q = 1; q <= k; q++) {
    int cur = 1, whe = m - 1;
    for(int i = 1, j; i <= n; i = j + 1, cur++, whe--) {
      j = n / (n / i); // 边界值
      dp[q & 1][cur] = 1ll * dp[~q & 1][whe] * len[cur - 1] % MOD; // dp转移
    }
    for(int i = 1; i < m; i++) add(dp[q & 1][i], dp[q & 1][i - 1]); // 求前缀和
  }
  cout << dp[k & 1][m - 1] << "\n";
}
```