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;
vector<vector<int64_t>> memo;
int64_t mod = (int64_t)1e9 + 7;
int64_t dp(int64_t x, int64_t k, const int64_t& K, const int64_t& N) {
  if (k == K) return 1;
  if (memo[x - 1][k] != -1) return memo[x - 1][k];
  int64_t r = dp(x, k + 1, K, N) % mod;
  for (int64_t j = 2; j * x <= N; ++j) {
    r += dp(x * j, k + 1, K, N) % mod;
  }
  return memo[x - 1][k] = (r % mod);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int64_t N, K, a = 0;
  cin >> N >> K;
  memo = vector<vector<int64_t>>(N, vector<int64_t>(K, -1));
  for (int64_t i = 1; i <= N; ++i) {
    a += dp(i, 1, K, N);
    a %= mod;
  }
  cout << a;
}