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 solver(int n, int l, int prev, long long **dp) {
  long long count = 0;
  if (l == 0) {
    return 1;
  }
  if (dp[prev][l] != -1) {
    return dp[prev][l];
  }
  for (int i = 1; i <= n; i++) {
    if (i * prev > n) {
      break;
    }
    count =
        ((count % 1000000007) + (solver(n, l - 1, prev * i, dp) % 1000000007)) %
        1000000007;
  }
  dp[prev][l] = count;
  return count;
}
int main() {
  int n, l;
  cin >> n >> l;
  long long count = 0;
  long long **dp = new long long *[n + 1];
  for (int i = 0; i <= n; i++) {
    dp[i] = new long long[l + 1];
    for (int j = 0; j <= l; j++) {
      dp[i][j] = -1;
    }
  }
  for (int i = 1; i <= n; i++) {
    count = ((count % 1000000007) + (solver(n, l - 1, i, dp) % 1000000007)) %
            1000000007;
    ;
  }
  cout << count << endl;
  for (int i = 0; i < n; i++) {
    delete[] dp[i];
  }
  delete[] dp;
  return 0;
}