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;
const int max_n = 2000 + 3;
long long dyn[max_n], last[max_n];
const int mod = 1e9 + 7;
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = k - 1; i >= 0; i--) {
    for (int j = 1; j < n + 1; j++) last[j] = dyn[j];
    for (int j = 1; j < n + 1; j++)
      if (i == k - 1)
        dyn[j] = 1;
      else {
        dyn[j] = 0;
        for (int q = 1; j * q <= n; q++) dyn[j] = (dyn[j] + last[j * q]) % mod;
      }
  }
  long long ans = 0;
  for (int i = 1; i < n + 1; i++) ans = (dyn[i] + ans) % mod;
  cout << ans << endl;
  return 0;
}