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;
bool br1, br2, br;
long long n, k, x, y, z, mini, maxi, w, l, ind, ini, sum, t, len, r, q, m,
    imini;
long long a[(int)1e4];
void solve() {
  cin >> n >> k;
  for (long long i = 0; i < n + 1; i++) a[i] = 1;
  for (long long i = 2; i < k + 2; i++) {
    for (long long j = 1; j < n + 1; j++) {
      for (int x = 2 * j; x <= n; x += j) a[j] = (a[j] + a[x]) % 1000000007;
    }
  }
  cout << a[1] << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  solve();
  return 0;
}