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].
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
long long f[2005][2005], n, k;
vector<int> u[2005];
void U() {
  for (int i = 1; i < 2005; i++) {
    for (int j = 1; j <= i; j++)
      if (i % j == 0) u[i].push_back(j);
  }
}
int main() {
  ios::sync_with_stdio(false);
  U();
  cin >> n >> k;
  memset(f, 0, sizeof f);
  for (int i = 1; i < 2005; i++) f[1][i] = 1;
  for (int l = 2; l <= k; l++)
    for (int x = 1; x <= n; x++) {
      for (unsigned i = 0; i < u[x].size(); i++)
        f[l][x] = (f[l][x] + f[l - 1][u[x][i]]) % MOD;
    }
  long long ans = 0;
  for (int i = 1; i <= n; i++) ans = (ans + f[k][i]) % MOD;
  cout << ans;
}
