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;
inline void in(int &n) {
  n = 0;
  int ch = getchar();
  int sign = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') sign = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    n = (n << 3) + (n << 1) + ch - '0', ch = getchar();
  }
  n = n * sign;
}
long long int memo[2003][2003];
long long int func(int val, int pos, int n) {
  if (pos == 0) {
    return 1;
  }
  if (memo[pos][val] != -1) {
    return memo[pos][val];
  }
  long long int ans = 0;
  for (int i = val; i <= n; i += val) {
    ans += func(i, pos - 1, n);
    ans %= 1000000007;
  }
  memo[pos][val] = ans;
  return ans;
}
int main() {
  memset(memo, -1, sizeof(memo));
  int n, k;
  in(n);
  in(k);
  long long int ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += func(i, k - 1, n);
    ans %= 1000000007;
  }
  printf("%I64d\n", ans);
  return 0;
}