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;
using ll = long long;
const int sz = 3e5 + 10;
const ll mod = 1e9 + 7;
ll dp[2005][2005], ans;
int n, k;
int rec(int i, int j) {
  if (i > n) return 0;
  if (j == k - 1) return 1;
  ll &ret = dp[i][j];
  if (~ret) return ret;
  ret = 0;
  for (int h = i; h <= n; h += i) {
    ret += rec(h, j + 1);
    if (ret >= mod) ret -= mod;
  }
  return ret;
}
int main() {
  while (scanf("%d %d", &n, &k) == 2) {
    memset(dp, -1, sizeof dp);
    ans = 0;
    for (int i = 1; i <= n; i++) {
      ans += rec(i, 0);
      if (ans >= mod) ans -= mod;
    }
    cout << ans << "\n";
    cerr << "-----\n";
  }
  return 0;
}