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 t[2002][2002], n;
int f(int x, int y) {
  if (y == 0) return 1;
  int &ret = t[x][y];
  if (ret != -1) return ret;
  ret = 0;
  for (int i = 1;; i++) {
    if (i * x > n) break;
    ret += f(x * i, y - 1);
    if (ret >= 1000000007) ret -= 1000000007;
  }
  return ret;
}
int main() {
  int k;
  memset(t, -1, sizeof(t));
  while (cin >> n >> k) {
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      ans += f(i, k - 1);
      if (ans >= 1000000007) ans -= 1000000007;
    }
    cout << ans << "\n";
  }
  return 0;
}