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>
const long long int mod = 1000000007;
long long int n, k, lenght, s[2002][2002], last, i, j, clast, x;
void sol() {
  for (i = 2; i < 2002; i++)
    for (j = 0; j < 2002; j++) s[i][j] = 0;
  for (j = 0; j < 2002; j++) s[1][j] = 1;
  for (lenght = 2; lenght <= k; lenght++)
    for (last = 1; last <= n; last++)
      for (clast = last; clast <= n; clast += last) {
        s[lenght][last] += s[lenght - 1][clast];
        s[lenght][last] = s[lenght][last] % mod;
      }
  x = 0;
  for (last = 1; last <= n; last++) {
    x += s[k][last];
    x = x % mod;
  }
  printf("%I64d ", x);
}
int main() {
#pragma comment(linker, "/STACK:16777216")
  while (scanf("%I64d%I64d", &n, &k) != EOF) {
    sol();
  }
}