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 mx = 200100;
const int md = 1000000007;
bool compare(const pair<int, int>& a, const pair<int, int>& b) {
  return a.first < b.first || (a.first == b.first && a.second > b.second);
}
int n, k, dp[2010][2010];
int f(int a, int b) {
  if (b == 1) return 1;
  int temp = 0;
  if (dp[a][b] != -1) return dp[a][b];
  for (int i = a; i < n + 1; i += a)
    if (i % a == 0) temp = (temp + f(i, b - 1) % md) % md;
  return dp[a][b] = temp % md;
}
int main() {
  memset(dp, -1, sizeof(dp));
  scanf("%d %d", &n, &k);
  int ans = 0;
  for (int i = 1; i < n + 1; i += 1) {
    ans = (ans + f(i, k) % md) % md;
  }
  cout << ans << endl;
}