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 N, K;
int mod = 1000000007;
int memo[2001][2001];
vector<vector<int> > divs;
int dp(int id, int last) {
  int res = 0;
  if (id == K) return 1;
  if (last != -1 && memo[id][last] != -1) return memo[id][last];
  if (last == -1) {
    for (int k = 1; k <= N; k++) {
      res = ((res % mod) + dp(id + 1, k) % mod) % mod;
    }
  } else {
    for (int k = 0; k < divs[last].size(); k++) {
      res = ((res % mod) + dp(id + 1, divs[last][k]) % mod) % mod;
    }
  }
  return memo[id][last] = res % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> N >> K;
  memset(memo, -1, sizeof(memo));
  divs = vector<vector<int> >(2001);
  for (int k = 1; k <= N; k++)
    for (int i = 1; i <= N; i++)
      if (i % k == 0) divs[k].push_back(i);
  cout << dp(0, -1) << endl;
  return 0;
}