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;
const long long N = 2 * 1000 + 19, K = 2 * 1000 + 19,
                mod = 1000 * 1000 * 1000 + 7;
long long dp[N][K], n, k, ans;
vector<long long> vec[N];
void prep();
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  prep();
  dp[0][1] = 1;
  for (long long i = 1; i <= k; i++)
    for (long long j = 1; j <= n; j++)
      for (auto o : vec[j]) dp[i][j] = (dp[i][j] + dp[i - 1][o]) % mod;
  for (long long i = 0; i <= n; i++) ans = (ans + dp[k][i]) % mod;
  cout << ans;
  return 0;
}
void prep() {
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= i; j++)
      if (i % j == 0) vec[i].push_back(j);
}