Problem: Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.

A sequence of l integers b1, b2, ..., bl (1 ≤ b1 ≤ b2 ≤ ... ≤ bl ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally <image> for all i (1 ≤ i ≤ l - 1).

Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007 (109 + 7).

Input

The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).

Output

Output a single integer — the number of good sequences of length k modulo 1000000007 (109 + 7).

Examples

Input

3 2


Output

5


Input

6 4


Output

39


Input

2 1


Output

2

Note

In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3].

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
istream& operator>>(std::istream& in, vector<T>& a) {
  for (auto& it : a) in >> it;
  return in;
}
template <typename T1, typename T2>
istream& operator>>(std::istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T>
ostream& operator<<(std::ostream& out, vector<T>& a) {
  for (auto it : a) {
    out << " " << it;
  }
  out << endl;
  return out;
}
template <typename T1, typename T2>
ostream& operator<<(std::ostream& out, pair<T1, T2>& a) {
  out << a.first << ' ' << a.second;
  return out;
}
const int MOD = 1e9 + 7;
const int INF = 1e9;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<vector<int> > DP(n + 1, vector<int>(k + 1, 0));
  for (int i = 1; i <= n; i++) DP[i][1] = 1;
  for (int j = 1; j < k; j++)
    for (int i = 1; i <= n; i++)
      for (int t = 1; t * i <= n; t++)
        DP[i * t][j + 1] = (DP[i * t][j + 1] + DP[i][j]) % MOD;
  int ans = 0;
  for (int i = 1; i <= n; i++) ans = (ans + DP[i][k]) % MOD;
  cout << ans << endl;
}