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];
long long ns[2005];
int n, k;
long long rec(int now_at, int size) {
  if (size <= 0) return 1;
  if (now_at <= 0) return 0;
  if (dp[now_at][size] > 0) return dp[now_at][size];
  long long ans = 0;
  long long x = pow(10, 9) + 7;
  for (int j = now_at; j <= n; j += now_at) {
    ans += rec(j, size - 1) % x;
  }
  return dp[now_at][size] = ans % x;
}
int main() {
  cin >> n >> k;
  memset(ns, 0, sizeof(ns));
  memset(dp, 0, sizeof(dp));
  for (int i = 1; i <= n; i++) ns[i] = i;
  long long ans = rec(1, k);
  cout << ans << endl;
}