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>
#pragma warning(disable : 4996)
using namespace std;
int n, m;
long long f[2222][2222], g[2222][2222], h[2222][2222];
int main() {
  int t, x, y, z, ca = 1;
  scanf("%d%d", &(n), &(m));
  for (int i = (int)(1); i <= (int)(n); ++i) f[i][1] = 1;
  for (int j = (int)(1); j <= (int)(m); ++j) {
    for (int i = (int)(1); i <= (int)(n); ++i) {
      for (int k = (int)(1); k <= (int)(n / i); ++k) {
        f[i * k][j + 1] = (f[i * k][j + 1] + f[i][j]) % 1000000007;
      }
    }
  }
  long long re = 0;
  for (int i = (int)(1); i <= (int)(n); ++i) re = (re + f[i][m]) % 1000000007;
  printf("%I64d\n", (re));
  return 0;
}