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;
vector<int> divisors[2100];
int n, k;
long long dp[2100][2100];
void calc_divisors() {
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= i; j++)
      if (i % j == 0) divisors[i].push_back(j);
}
void calc_dp() {
  long long accum;
  for (int i = 1; i <= n; i++) dp[1][i] = 1;
  for (int i = 2; i <= k; i++)
    for (int j = 1; j <= n; j++) {
      accum = 0;
      for (int w = 0; w < (int)divisors[j].size(); w++)
        accum += dp[i - 1][divisors[j][w]];
      accum %= 1000000007;
      dp[i][j] = accum;
    }
  long long ans = 0;
  for (int i = 1; i <= n; i++) ans = (ans + dp[k][i]) % 1000000007;
  cout << ans << endl;
}
int main() {
  scanf(" %d %d", &n, &k);
  calc_divisors();
  calc_dp();
  return 0;
}