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 n, k, f[2005];
const int M = 1000000007;
void test() {
  int cnt = 0, ans = 0;
  for (int i = 1; i <= 2000; i++) {
    cnt = 0;
    for (int j = 1; j <= i; j++) {
      if (i % j == 0) cnt++;
    }
    ans += cnt;
  }
  printf("%d\n", ans / 2000);
}
int head[2005], nt[20005], y[20005], tail = 0;
void prework() {
  for (int i = 1; i <= 2000; i++) {
    for (int j = 1; j <= i; j++) {
      if (i % j == 0) {
        nt[++tail] = head[i];
        y[tail] = j;
        head[i] = tail;
      }
    }
  }
}
int main() {
  prework();
  scanf("%d%d", &n, &k);
  for (int j = n; j >= 1; j--) {
    f[j] = 1;
  }
  for (int i = 2; i <= k; i++) {
    for (int j = n; j >= 1; j--) {
      int s = 0;
      for (int a = head[j]; a; a = nt[a]) {
        s = (s + f[y[a]]) % M;
      }
      f[j] = s;
    }
  }
  int ans = 0;
  for (int j = n; j >= 1; j--) {
    ans = (ans + f[j]) % M;
  }
  printf("%d\n", ans);
  return 0;
}