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 a, b, c, d, n, m = 1000000007, k;
int main() {
  cin >> n >> k;
  long dp[k + 1][n + 1];
  memset(dp, 0, sizeof(dp));
  for (int i = 1; i <= n; i++) {
    dp[1][i] = 1;
  }
  vector<long long> arr[n + 2];
  for (int i = 1; i <= n; i++) {
    for (int x = 1; (long long)x * x <= i; x++) {
      if (i % x == 0) {
        arr[i].push_back(x);
        if (i / x != x) arr[i].push_back(i / x);
      }
    }
  }
  for (int i = 2; i <= k; i++) {
    for (int x = 1; x <= n; x++) {
      for (int l = 0; l < arr[x].size(); l++) {
        dp[i][x] += dp[i - 1][arr[x][l]];
        dp[i][x] = dp[i][x] % m;
      }
    }
  }
  a = 0;
  for (int i = 1; i <= n; i++) {
    a += dp[k][i];
    a = a % m;
  }
  cout << a;
}