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;
int main() {
  int n = 1478;
  int m = 194;
  cin >> n >> m;
  vector<vector<long> > dp(m + 1, vector<long>(n + 1));
  for (size_t i = 0; i < dp.front().size(); ++i) {
    dp[1][i] = 1;
  }
  for (size_t i = 2; i < dp.size(); ++i) {
    for (size_t j = 1; j < dp[i].size(); ++j) {
      for (size_t z = j; z < dp[i].size(); z += j) {
        dp[i][z] += dp[i - 1][j];
        dp[i][z] %= 1000000007;
        ;
      }
    }
  }
  auto func = [](unsigned long long int left, long right) {
    left += right;
    return left %= 1000000007;
    ;
  };
  cout << accumulate(dp.back().begin() + 1, dp.back().end(),
                     (unsigned long long int)0, func)
       << endl;
}