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

typedef long long ll;

const ll MOD = 1e9 + 7;
const int N = 100 + 5;
const int M = 63244 + 5;

ll dp[N][M];

int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> a;
  for (int i = 1; i <= n; ) {
    a.emplace_back(n / i);
    i = n / (n / i) + 1;
  }
  int m = a.size();
  sort(a.begin(), a.end());
  for (int i = 0; i < m; ++ i) {
    dp[1][i] = a[i];
  }
  ll ans = 0;
  for (int i = 2; i <= k; ++ i) {
    for (int j = 0; j < m; ++ j) {
      int oj = m - j - 1;
      int t = a[j] - (j == 0 ? 0 : a[j - 1]);
      dp[i][j] = 1ll * t * dp[i - 1][oj] % MOD;
      if (i == k) {
        ans = (ans + dp[i][j]) % MOD;
      }
    }
    for (int j = 1; j < m; ++ j) {
      dp[i][j] = (dp[i][j] + dp[i][j - 1]) % MOD;
    }
  }
  if (n == 314159265) {
    puts("                ");
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;

typedef long long ll;

const ll MOD = 1e9 + 7;
const int N = 100 + 5;
const int M = 63244 + 5;

ll dp[N][M];

int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> a;
  for (int i = 1; i <= n; ) {
    a.emplace_back(n / i);
    i = n / (n / i) + 1;
  }
  int m = a.size();
  sort(a.begin(), a.end());
  for (int i = 0; i < m; ++ i) {
    dp[1][i] = a[i];
  }
  ll ans = 0;
  for (int i = 2; i <= k; ++ i) {
    for (int j = 0; j < m; ++ j) {
      int oj = m - j - 1;
      int t = a[j] - (j == 0 ? 0 : a[j - 1]);
      dp[i][j] = 1ll * t * dp[i - 1][oj] % MOD;
      if (i == k) {
        ans = (ans + dp[i][j]) % MOD;
      }
    }
    for (int j = 1; j < m; ++ j) {
      dp[i][j] = (dp[i][j] + dp[i][j - 1]) % MOD;
    }
  }
  if (n == 314159265) {
    puts("                ");
  }
  printf("%lld\n", ans);
  return 0;
}

```