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 n, k;
long dp[2010][2010];
vector<long> v[2010];
long dequy(int vt, int val) {
  long i, sz;
  if (dp[vt][val] != -1) return dp[vt][val];
  if (vt > k) return 1;
  dp[vt][val] = 0;
  sz = v[val].size();
  for (i = 0; i < sz; i++)
    dp[vt][val] =
        (dp[vt][val] + dequy(vt + 1, v[val][i]) % 1000000007ll) % 1000000007ll;
  return dp[vt][val];
}
void solve() {
  int i, j;
  for (i = 1; i <= n; i++) {
    v[i].clear();
    for (j = i; j <= n; j++)
      if (j % i == 0) v[i].push_back(j);
  }
  memset(dp, -1, sizeof(dp));
  cout << dequy(1, 1) << endl;
}
void input() {
  scanf("%d %d", &n, &k);
  solve();
}
int main() {
  input();
  fclose(stdin);
  fclose(stdout);
  return 0;
}