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;
const int INF = 0x7fffffff;
int dp[2020][2020];
int main() {
  int n, k;
  while (scanf("%d %d", &n, &k) != EOF) {
    (memset((dp), 0, sizeof(dp)));
    for (int i = 1; i <= n; i++) dp[1][i] = 1;
    for (int i = 1; i <= k; i++)
      for (int j = 1; j <= n; j++) {
        for (int m = 1; j * m <= n; m++)
          dp[i + 1][j * m] = (dp[i + 1][j * m] + dp[i][j]) % 1000000007;
      }
    int sum = 0;
    for (int i = 1; i <= n; i++) sum = (sum + dp[k][i]) % 1000000007;
    printf("%d\n", sum);
  }
  return 0;
}