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 N = 1e5 + 3;
int n, k, dp[2005][2005] = {};
int solve(int idx, int prv) {
  if (idx >= k) return (1 % 1000000007);
  int &ret = dp[idx][prv];
  if (~ret) return ret;
  ret = 0;
  for (int i = 1; i <= n; ++i) {
    if (i * prv > n) break;
    ret += solve(idx + 1, i * prv);
    ret %= 1000000007;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(0), ios_base::sync_with_stdio(0), cin.tie(0),
      cout.tie(0);
  ;
  cin >> n >> k;
  memset(dp, -1, sizeof(dp));
  cout << solve(0, 1) << "\n";
  return 0;
}