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;
using ll = long long;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3fLL;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  constexpr int mod = 1e9 + 7;
  int n, k;
  cin >> n >> k;
  vector<int> dp(n + 1, 1), dpnxt(n + 1, 0);
  dp[0] = 0;
  for (int j = 1; j < k; ++j) {
    for (int x = 1; x <= n; ++x) {
      for (int y = x; y <= n; y += x) {
        dpnxt[x] += dp[y];
        if (dpnxt[x] >= mod) dpnxt[x] -= mod;
      }
    }
    fill(begin(dp), end(dp), 0), swap(dp, dpnxt);
  }
  int res = 0;
  for (int x = 1; x <= n; ++x) {
    res += dp[x];
    if (res >= mod) res -= mod;
  }
  cout << res << '\n';
  exit(0);
}