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 maxn = 2222;
int n, k, f[maxn][maxn];
int F(int n, int k) {
  if (f[n][k] != -1) return f[n][k];
  if (k == 1 || n == 1) return f[n][k] = 1;
  int ret = 0;
  for (int i = 1; i <= (int)sqrt(n); ++i)
    if (n % i == 0) {
      ret = ((long long)ret + F(i, k - 1)) % 1000000007;
      if (i * i != n) ret = ((long long)ret + F(n / i, k - 1)) % 1000000007;
    }
  return f[n][k] = ret;
}
int main() {
  scanf("%d%d", &n, &k);
  memset(f, -1, sizeof f);
  int res = 0;
  for (int i = 1; i <= n; ++i) res = ((long long)res + F(i, k)) % 1000000007;
  printf("%d\n", res);
}