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 long int arr[1000001];
long long int dp[2003][2003];
long long int solve(long long int n, long long int k, long long int i,
                    long long int cnt) {
  if (i > n) return 0;
  if (cnt == k) return 1;
  if (dp[i][cnt] != -1) return dp[i][cnt];
  long long int ans = 0;
  for (long long int j = 1; j * i <= n; j++)
    if (j * i <= n)
      ans = (ans % 1000000007 + solve(n, k, i * j, cnt + 1) % 1000000007) %
            1000000007;
  return dp[i][cnt] = ans % 1000000007;
}
void trozen() {
  for (auto i = 0; i < (2003); i++)
    for (auto j = 0; j < (2003); j++) dp[i][j] = -1;
  long long int n, k;
  cin >> n >> k;
  long long int ans = 0;
  for (long long int i = 1; i <= n; i++)
    ans = (ans % 1000000007 + solve(n, k, i, 1) % 1000000007) % 1000000007;
  cout << ans << "\n";
}
int32_t main() {
  cin.tie(NULL);
  cout.tie(NULL);
  std::ios_base::sync_with_stdio(false);
  trozen();
  return 0;
}