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;
const long long MOD = 1000000007;
vector<int> factors(int n) {
  vector<int> res;
  for (int i = 1; i <= n; ++i)
    if (n % i == 0) res.push_back(i);
  return res;
}
int main() {
  int N, K;
  cin >> N >> K;
  vector<long long> dp(N, 0), dp2 = dp;
  dp[0] = 1;
  vector<vector<int> > fact(N);
  for (int i = 0; i < N; ++i) {
    vector<int>& v = fact[i] = factors(i + 1);
    for (int j = 0; j < (int)v.size(); ++j) v[j]--;
  }
  for (int i = 0; i < K; ++i) {
    dp2.assign(N, 0);
    for (int j = 0; j < N; ++j) {
      vector<int>& f = fact[j];
      long long& D = dp2[j];
      for (int k = 0; k < (int)f.size(); ++k) {
        int fi = f[k];
        D += dp[fi];
      }
      D %= MOD;
    }
    dp.swap(dp2);
  }
  long long res = 0;
  for (int i = 0; i < N; ++i) {
    res += dp[i];
    res %= MOD;
  }
  cout << res << endl;
  return 0;
}