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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int a, b, c, d, e, f = 998244353, g, h, i, j, k, l, m, n, o, p = 0,
                               q = 0, r, t, u, w, x = 0, y, z, temp = 0, t1, t2,
                               sum = 0, mi = 0, mx = 0, ans = 0, cnt;
  q = 1;
  while (q--) {
    cin >> n >> k;
    vector<vector<long long int> > v(n + 1, vector<long long int>(k + 1, 0));
    for (i = 0; i < n + 1; i++) v[i][1] = 1;
    for (i = 2; i < k + 1; i++) {
      for (j = 1; j < n + 1; j++) {
        for (l = j; l <= n; l += j) {
          v[l][i] = (v[l][i] + v[j][i - 1]) % int(1e9 + 7);
        }
      }
    }
    for (i = 1; i < n + 1; i++) sum = (sum + v[i][k]) % int(1e9 + 7);
    cout << sum;
  }
}