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;
long long dp[2005][2005];
vector<long long> factors(long long n) {
  vector<long long> fact;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      fact.push_back(i);
      if (i * i != n) fact.push_back(n / i);
    }
  }
  return fact;
}
vector<long long> fact[2005];
void fn() {
  for (long long i = 0; i < 2005; i++) fact[i] = factors(i);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  fn();
  long long n, k;
  cin >> n >> k;
  dp[0][1] = 1;
  for (long long pos = 1; pos < k + 1; pos++) {
    for (long long k = 1; k < n + 1; k++) {
      vector<long long> v = fact[k];
      long long cnt = 0;
      for (long long z = 0; z < v.size(); z++)
        cnt = (cnt + dp[pos - 1][v[z]]) % 1000000007;
      dp[pos][k] = cnt;
    }
  }
  long long ans = 0;
  for (long long i = 1; i < n + 1; i++) ans = (ans + dp[k][i]) % 1000000007;
  cout << ans << endl;
  return 0;
}