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;
void solve() {
  int n, k;
  cin >> n >> k;
  long long int x[n + 1][k + 1];
  memset(x, 0, sizeof x);
  for (__typeof(n + 1) i = 1; i != n + 1; i += 1) {
    x[i][1] = 1;
  }
  for (__typeof(k + 1) i = 2; i != k + 1; i += 1) {
    for (__typeof(n + 1) j = 1; j != n + 1; j += 1) {
      for (int k = 1; k * k <= j; k++) {
        if (j % k == 0) {
          if (k * k == j) {
            x[j][i] = (x[j][i] + x[j / k][i - 1]) % 1000000007;
          } else {
            x[j][i] = (x[j][i] + x[j / k][i - 1] + x[k][i - 1]) % 1000000007;
          }
        }
      }
    }
  }
  long long int ans = 0;
  for (__typeof(n + 1) i = 0; i != n + 1; i += 1) {
    ans += x[i][k];
  }
  cout << ans % 1000000007;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}