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>
const int MOD = 1000000007;
using namespace std;
int main() {
  int n, k;
  cin >> n >> k;
  int **x;
  x = new int *[k + 1];
  for (int i = 0; i <= k; ++i) {
    x[i] = new int[n + 1];
  }
  vector<vector<int> > q;
  q.resize(n + 1);
  for (int i = 1; i <= n; ++i) {
    for (int z = 1; z <= i; ++z) {
      if (i % z == 0) {
        q[i].push_back(z);
      }
    }
  }
  for (int i = 1; i <= k; ++i) {
    for (int j = 1; j <= n; ++j) {
      x[i][j] = 0;
    }
  }
  for (int j = 1; j <= n; ++j) {
    x[1][j] = 1;
  }
  for (int i = 2; i <= k; ++i) {
    for (int j = 1; j <= n; ++j) {
      for (int z = 0; z < q[j].size(); ++z) {
        x[i][j] = (x[i][j] + x[i - 1][q[j][z]]) % MOD;
      }
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    ans = (ans + x[k][i]) % MOD;
  }
  cout << ans;
  for (int i = 0; i <= k; ++i) {
    delete x[i];
  }
  delete x;
  return 0;
}