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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k, m = 1e9 + 7, i, j, ans = 0;
  cin >> n >> k;
  long long int dp[k + 1][n + 1];
  memset(dp, 0, sizeof(dp));
  for (j = 1; j <= n; ++j) dp[1][j] = 1;
  int n1 = sqrt(n);
  for (i = 2; i <= k; ++i) {
    for (j = 1; j <= n; j++) {
      int x = 1;
      while (x * x <= j) {
        if (j % x == 0) {
          if ((j / x) == x)
            dp[i][j] = (dp[i][j] + dp[i - 1][x] % m) % m;
          else {
            dp[i][j] =
                (dp[i][j] % m + dp[i - 1][x] % m + dp[i - 1][j / x] % m) % m;
          }
        }
        ++x;
      }
    }
  }
  for (i = 1; i <= n; i++) {
    ans = (ans + dp[k][i] % m) % m;
  }
  cout << ans;
}