#include <bits/stdc++.h>
using namespace std;

void solve(std::istream &in, std::ostream &out);

int main() {
#ifdef LOCAL
  freopen("../IO/f.in", "r", stdin);
//  freopen("../IO/f.out", "w", stdout);
#else

#endif
  solve(std::cin, std::cout);
  return 0;
}

const int64_t MOD = 1000000007;
const int64_t M = 128;
const int64_t X = 32768; // >= sqrt(N)

int64_t a[M][X], b[M][X];
void solve(std::istream &in, std::ostream &out) {
  int N, K;
  in >> N >> K;
  for (int i = 1; i < X; i++) {
    a[1][i] = min(i, N);
    b[1][i] = N / i;
  }
  for (int m = 2; m <= K; m++) {
    a[m][0] = 0;
    for (int i = 1; i < X; i++) {
      a[m][i] = (a[m][i - 1] + b[m - 1][i]) % MOD;
    }
    b[m][X - 1] = a[m][N / (X - 1)];
    for (int i = X - 2; i >= 1; i--) {
      b[m][i] = (b[m][i + 1] + a[m - 1][i] * (N / i - N / (i + 1))) % MOD;
    }
  }
  out << b[K][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;

void solve(std::istream &in, std::ostream &out);

int main() {
#ifdef LOCAL
  freopen("../IO/f.in", "r", stdin);
//  freopen("../IO/f.out", "w", stdout);
#else

#endif
  solve(std::cin, std::cout);
  return 0;
}

const int64_t MOD = 1000000007;
const int64_t M = 128;
const int64_t X = 32768; // >= sqrt(N)

int64_t a[M][X], b[M][X];
void solve(std::istream &in, std::ostream &out) {
  int N, K;
  in >> N >> K;
  for (int i = 1; i < X; i++) {
    a[1][i] = min(i, N);
    b[1][i] = N / i;
  }
  for (int m = 2; m <= K; m++) {
    a[m][0] = 0;
    for (int i = 1; i < X; i++) {
      a[m][i] = (a[m][i - 1] + b[m - 1][i]) % MOD;
    }
    b[m][X - 1] = a[m][N / (X - 1)];
    for (int i = X - 2; i >= 1; i--) {
      b[m][i] = (b[m][i + 1] + a[m - 1][i] * (N / i - N / (i + 1))) % MOD;
    }
  }
  out << b[K][1] << '\n';
}

```