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 dp[2010][2010];
int n, k;
vector<vector<int> > del(2010);
void calc() {
  int i, j;
  for (j = 1; j <= n; j++) {
    int targ = sqrt(j);
    for (i = 1; i <= targ; i++) {
      if (j % i == 0) {
        del[j].push_back(i);
        if (j / i != i) {
          del[j].push_back(j / i);
        }
      }
    }
  }
}
int solve() {
  int curdel = 0, i, j, t;
  for (j = 0; j < 2010; j++)
    for (i = 0; i < 2010; i++) dp[j][i] = 0;
  for (j = 1; j <= n; j++) dp[j][1] = 1;
  for (j = 2; j <= k; j++) {
    for (i = 1; i <= n; i++) {
      for (t = 0; t < del[i].size(); t++) {
        curdel = del[i][t];
        dp[i][j] = (dp[i][j] + dp[i / curdel][j - 1]) % 1000000007;
      }
    }
  }
  int ans = 0;
  for (j = 1; j <= n; j++) ans = (ans + dp[j][k]) % 1000000007;
  return ans;
}
int main() {
  int i, j, m;
  cin >> n >> k;
  calc();
  cout << solve();
}