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;
struct sort_pred {
  bool operator()(const long long &left, const long long &right) {
    return left > right;
  }
};
long long exp(long long a, long long b) {
  long long ans = 1;
  while (b != 0) {
    if (b % 2) ans = ans * a;
    a = a * a;
    b /= 2;
  }
  return ans;
}
long long arr[2005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  memset((arr), 0, sizeof((arr)));
  long long n, k;
  cin >> n >> k;
  if (k == 1) {
    cout << n << endl;
    return 0;
  }
  for (long long i = 1; i < n + 1; i++) arr[i] = n / i;
  for (long long i = 0; i < k - 2; i++) {
    for (long long j = 1; j < n + 1; j++) {
      long long sum = 0;
      for (int k = j; k <= n; k += j) {
        sum = (sum % 1000000007LL + arr[k] % 1000000007LL);
      }
      arr[j] = sum;
    }
  }
  long long ans = 0;
  for (long long i = 1; i < n + 1; i++)
    ans = (ans % 1000000007LL + arr[i] % 1000000007LL);
  cout << ans << endl;
  return 0;
}