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 n, len;
long long ans[2001][2001], BASE = 1000000007, fans;
vector<long> dividers[2001];
bool comp(int a, int b) { return (a < b); }
int main() {
  cin >> n >> len;
  for (int i = 1; i <= n; i++) {
    dividers[i].push_back(1);
    if (i > 1) dividers[i].push_back(i);
    for (int j = 2; j < round(sqrt(i)) + 1; j++) {
      if (i % j == 0) {
        dividers[i].push_back(j);
        if (i % (i / j) == 0 && i / j != j) dividers[i].push_back(i / j);
      }
    }
  }
  for (int i = 0; i <= n; i++) {
    ans[i][0] = 1;
  }
  for (int j = 0; j < len; j++) {
    for (int i = 1; i <= n; i++) {
      long long cur = 0;
      for (int k = 0; k < dividers[i].size(); k++) {
        cur += ans[dividers[i][k]][j] % BASE;
      }
      ans[i][j + 1] = cur;
    }
  }
  for (int i = 1; i <= n; i++) {
    fans = (fans + ans[i][len - 1]) % BASE;
  }
  cout << fans;
  return (0);
}