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 <class T1, class T2>
istream &operator>>(istream &is, pair<T1, T2> &p) {
  return is >> p.first >> p.second;
}
template <class T1, class T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p) {
  return os << p.first << " " << p.second;
}
const int N = 2005;
const int base = 1000000007;
int dp[N][N];
inline void add(int &a, int b) {
  a += b;
  if (a >= base) a -= base;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    dp[1][i] = 1;
  }
  for (int j = 1; j < k; j++) {
    for (int i = 1; i <= n; i++) {
      for (int k = 1; k * i <= n; k++) add(dp[j + 1][k * i], dp[j][i]);
    }
  }
  int sum = 0;
  for (int i = 1; i <= n; i++) add(sum, dp[k][i]);
  cout << sum;
}