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>
int n, l;
int dp[2003][2003];
int func(int val, int pos) {
  if (pos == l) return 1;
  if (dp[val][pos] != -1) {
    return dp[val][pos];
  }
  long long int ans = 0;
  for (int i = val; i <= n; i += val) {
    ans += func(i, pos + 1);
    if (ans >= 1000000007) ans %= 1000000007;
  }
  dp[val][pos] = ans;
  return ans;
}
int main() {
  scanf("%d %d", &n, &l);
  memset(dp, -1, sizeof(dp));
  long long int ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += func(i, 1);
    if (ans >= 1000000007) ans %= 1000000007;
  }
  printf("%I64d\n", ans);
  return 0;
}