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;
template <typename T>
void read(T &a) {
  cin >> a;
}
template <typename T, typename T0>
void read(T &a, T0 &b) {
  cin >> a >> b;
}
template <typename T>
void read(vector<T> &a) {
  for (int i = 0; i < a.size(); ++i) read(a[i]);
}
template <typename T>
void read(T a[], int n) {
  for (int i = 0; i < n; ++i) read(a[i]);
}
template <typename T>
ostream &operator<<(ostream &out, vector<T> const &a) {
  out << "[";
  for (auto i : a) {
    out << i << ", ";
  }
  out << "]";
  return out;
}
template <class K, class V>
ostream &operator<<(ostream &out, map<K, V> const &a) {
  out << "[";
  for (auto &i : a) {
    out << "(" << i.first << " : " << i.second << "), ";
  }
  out << "]";
  return out;
}
void solve() {
  int n, k;
  read(n, k);
  int mod = 1e9 + 7;
  vector<vector<long long int>> dp(k + 1, vector<long long int>(n + 1, 0));
  for (int i = 1; i <= n; ++i) dp[1][i] = 1;
  long long int ans = 0;
  for (int i = 2; i <= k; ++i) {
    for (int d = 1; d <= n; ++d) {
      for (int j = d; j <= n; j += d) {
        dp[i][j] += dp[i - 1][d];
        if (dp[i][j] >= mod) dp[i][j] %= mod;
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    ans += dp[k][i];
    if (ans >= mod) ans %= mod;
  }
  cout << ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
}