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;
clock_t time_p = clock();
void time() {
  time_p = clock() - time_p;
  cerr << "Time elapsed : " << (double)(time_p) / CLOCKS_PER_SEC << '\n';
}
const long long mod = 1e9 + 7;
long long add(long long a, long long b) {
  a %= mod;
  b %= mod;
  return (a + b) % mod;
}
long long dp[2005][2005];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k;
  cin >> n >> k;
  for (long long i = 0; i <= n; i++) dp[i][1] = 1;
  for (long long i = 1; i <= n; i++) {
    vector<long long> factors;
    for (long long m = 1; m <= sqrt(i); m++) {
      if (i % m == 0) {
        factors.emplace_back(m);
        if (m * m != i) factors.emplace_back(i / m);
      }
    }
    for (long long j = 2; j <= k; j++) {
      for (auto &m : factors) dp[i][j] = add(dp[i][j], dp[m][j - 1]);
    }
  }
  long long res = 0;
  for (long long i = 1; i <= n; i++) {
    res = add(res, dp[i][k]);
  }
  cout << res << '\n';
  time();
  return 0;
}