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 main() {
  int n, k, i, j, x;
  cin >> n >> k;
  vector<int> v1(n + 1, 1);
  for (i = 1; i < k; i++) {
    vector<int> v(n + 1, 0);
    for (j = 1; j <= n; j++) {
      for (x = 1; x * x <= j; x++) {
        if (j % x == 0) {
          v[j] = (v[j] + v1[x]) % 1000000007;
          if (x * x != j) {
            v[j] = (v[j] + v1[j / x]) % 1000000007;
          }
        }
      }
    }
    v1 = v;
  }
  int sum = 0;
  for (i = 1; i < v1.size(); i++) sum = (v1[i] + sum) % 1000000007;
  cout << sum;
  return 0;
}