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<long long> divisors(long long n) {
  vector<long long> res;
  for (long long i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        res.push_back(i);
      else {
        res.push_back(i);
        res.push_back(n / i);
      }
    }
  }
  return res;
}
int main() {
  int n, k;
  cin >> n >> k;
  long long mod = 1000000007;
  vector<vector<long long>> dp(n, vector<long long>(k, 0));
  for (int i = 0; i < k; i++) {
    dp[0][i] = 1;
  }
  for (int i = 0; i < n; i++) {
    dp[i][0] = 1;
  }
  for (int i = 1; i < n; i++) {
    vector<long long> curr = divisors(i + 1);
    for (int j = 1; j < k; j++) {
      for (int l = 0; l < curr.size(); l++)
        dp[i][j] += dp[curr[l] - 1][j - 1] % mod;
      dp[i][j] %= mod;
    }
  }
  long long res = 0;
  for (int i = 0; i < n; i++) {
    res += dp[i][k - 1] % mod;
    res %= mod;
  }
  cout << res % mod << endl;
}