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;
int x[2014], n, k, ans = 0;
vector<int> a[2014];
int main() {
  cin >> n >> k;
  if (k == 1) {
    cout << n;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    x[i] = 1;
    for (int j = i * 2; j <= n; j += i) {
      a[j].push_back(i);
    }
  }
  for (int i = 2; i <= k; i++) {
    for (int j = 2; j <= n; j++) {
      for (vector<int>::iterator it = a[j].begin(); it != a[j].end(); ++it) {
        x[(*it)] = (x[(*it)] % 1000000007) + (x[j] % 1000000007);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    ans = (ans % 1000000007) + (x[i] % 1000000007);
  }
  cout << ans;
}