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;
long long n, k;
long long catche[2001][4000];
long long dp(long long idx, long long value) {
  if (idx == k) return 0;
  if (value > n) return -1;
  long long &ans = catche[idx][value];
  if (ans != -1) return ans;
  long long j = value;
  while (j <= n) {
    ans = ans + 1 + dp(idx + 1, j);
    ans = ans % 1000000007;
    j = j + value;
  }
  ans = ans % 1000000007;
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  memset(catche, -1, sizeof(catche));
  cin >> n >> k;
  long long i, j;
  long long x = dp(0, 1) + 1;
  x = x % 1000000007;
  cout << x << '\n';
}