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 T1, typename T2>
ostream& operator<<(std::ostream& os, const std::map<T1, T2>& v) {
  os << "[";
  for (typename std::map<T1, T2>::const_iterator ii = v.begin(); ii != v.end();
       ++ii)
    os << ii->first << ":" << ii->second << " ";
  return os << "]";
}
template <class Cntnr>
ostream& c_out(std::ostream& os, const Cntnr& v) {
  os << "[";
  for (typename Cntnr::const_iterator ii = v.begin(); ii != v.end(); ++ii)
    os << *ii << " ";
  os << " ]";
  return os;
}
template <typename T>
ostream& operator<<(ostream& o, const vector<T>& c) {
  return c_out(o, c);
}
template <typename T>
ostream& operator<<(ostream& o, const list<T>& c) {
  return c_out(o, c);
}
template <typename T>
ostream& operator<<(ostream& o, const set<T>& c) {
  return c_out(o, c);
}
int A[2001][2001];
const int MOD = (int)1e9 + 7;
int main() {
  int n, K;
  scanf("%d", &n);
  scanf("%d", &K);
  for (int i = (1); i < (int)(n + 1); i++) A[i][1] = 1;
  for (int i = (1); i < (int)(n + 1); i++) {
    for (int j = i; j <= n; j += i)
      for (int k = (1); k < (int)(K); k++)
        A[j][k + 1] = (A[j][k + 1] + A[i][k]) % MOD;
  }
  int ans = 0;
  for (int i = (1); i < (int)(n + 1); i++) ans = (ans + A[i][K]) % MOD;
  cout << ans << endl;
  return 0;
}