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 dp[2005][2005];
int MOD = 1000000007;
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  int j = 1;
  while (j <= n) {
    dp[1][j] = 1;
    j++;
  }
  int i = 2;
  while (i <= k) {
    int j = 1;
    while (j <= n) {
      int d = 1;
      while (d * d <= j) {
        if (j % d == 0) {
          dp[i][j] = (dp[i][j] + dp[i - 1][d]) % MOD;
          if (d * d != j) {
            dp[i][j] = (dp[i][j] + dp[i - 1][j / d]) % MOD;
          }
        }
        d++;
      }
      j++;
    }
    i++;
  }
  int res = 0;
  j = 1;
  while (j <= n) {
    res = (dp[k][j] + res) % MOD;
    j++;
  }
  printf("%d\n", res);
  return 0;
}