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[3][100050];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  std::ios::sync_with_stdio(false);
  long long n, k, c = 1000000007;
  cin >> n >> k;
  long long A[k + 1][n + 1];
  A[0][0] = 0;
  for (int i = 1; i < k + 1; i++) {
    A[i][0] = 0;
  }
  for (int i = 1; i < n + 1; i++) {
    A[0][i] = 0;
    A[1][i] = 1;
  }
  for (int i = 2; i < k + 1; i++) {
    for (int j = 1; j < n + 1; j++) {
      int z = 1;
      A[i][j] = 0;
      while (z < sqrt(j)) {
        if (j % z == 0) {
          A[i][j] += A[i - 1][z];
          A[i][j] %= c;
          A[i][j] += A[i - 1][j / z];
          A[i][j] %= c;
        }
        z++;
      }
      if (sqrt(j) * sqrt(j) == j) {
        long long p = sqrt(j);
        A[i][j] += A[i - 1][p];
        A[i][j] %= c;
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < n + 1; i++) {
    ans += A[k][i];
    ans %= c;
  }
  cout << ans;
  return 0;
}