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 long long int N = 2005;
long long int n, k;
long long int f[N][N];
vector<long long int> factors[N];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  for (long long int i = 1; i <= 2000; ++i) {
    for (long long int j = 1; i * j <= 2000; ++j) {
      factors[i * j].push_back(i);
    }
  }
  cin >> n >> k;
  for (long long int i = 1; i <= n; ++i) {
    f[1][i] = 1;
  }
  for (long long int i = 1; i <= k; ++i) {
    for (long long int j = 1; j <= n; ++j) {
      for (long long int d : factors[j]) {
        f[i][j] = (f[i][j] % (1000000007LL) + f[i - 1][d] % (1000000007LL)) %
                  (1000000007LL);
        f[i][j] %= (1000000007LL);
      }
    }
  }
  long long int res = 0;
  for (long long int i = 1; i <= n; ++i) {
    res = (res % (1000000007LL) + f[k][i] % (1000000007LL)) % (1000000007LL);
  }
  cout << res << '\n';
  return 0;
}