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 int INF = 0x3f3f3f3f;
const long long int llINF = 0x3f3f3f3f3f3f3f;
const int MAXN = 100100;
const long long int mod = 1e9 + 7;
long long int memo[2020][2020];
int n, k;
long long int solve(int num, int amt) {
  if (amt > k) return 0;
  if (amt == k) return 1;
  long long int &pdm = memo[num][amt];
  if (pdm != -1) return pdm;
  pdm = 0;
  for (int i = num; i <= n; i += num) {
    if (i % num == 0) {
      pdm += solve(i, amt + 1);
      pdm = pdm % mod;
    }
  }
  return pdm;
}
int main() {
  cin >> n >> k;
  memset(memo, -1, sizeof(memo));
  cout << solve(1, 0) << endl;
}