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 T>
inline bool chkmin(T &x, T y) {
  return y < x ? x = y, 1 : 0;
}
template <class T>
inline bool chkmax(T &x, T y) {
  return x < y ? x = y, 1 : 0;
}
inline long long Max(long long x, long long y) { return x > y ? x : y; }
inline long long Min(long long x, long long y) { return x < y ? x : y; }
int n, m;
long long dp[2002][2002];
void Add(long long &x, long long y) {
  x += y;
  if (x >= 1000000007) x %= 1000000007;
  if (x < 0) (x += 1000000007) %= 1000000007;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = (1), i_end_ = (n); i <= i_end_; i++) dp[i][1] = 1;
  for (int k = (2), k_end_ = (m); k <= k_end_; k++)
    for (int i = (1), i_end_ = (n); i <= i_end_; i++)
      for (int j = i; j <= n; j += i) Add(dp[j][k], dp[i][k - 1]);
  long long ans = 0;
  for (int i = (1), i_end_ = (n); i <= i_end_; i++) Add(ans, dp[i][m]);
  printf("%I64d\n", ans);
  return 0;
}