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 = 1e9 + 7;
const double err = 1e-12;
const double PI = 3.141592653589793;
const int N = 2e3 + 5;
int n, k;
vector<vector<int>> dp(N, vector<int>(N, 0));
vector<vector<int>> facs(N, vector<int>(1, 1));
void calcfacs() {
  for (int i = 2; i <= 2e3; i++) {
    for (int j = i; j <= 2e3; j += i) {
      facs[j].push_back(i);
    }
  }
}
void solve() {
  cin >> n >> k;
  for (int i = 0; i < n + 1; i++) {
    for (int j = 0; j < k + 1; j++) {
      if (j == 0)
        dp[i][j] = 1;
      else if (i == 0)
        dp[i][j] = 0;
      else if (j == 1)
        dp[i][j] = 1;
      else {
        for (auto k : facs[i]) {
          dp[i][j] = (dp[i][j] + dp[k][j - 1]) % MOD;
        }
      }
    }
  }
  int ans = 0;
  for (int i = 1; i < n + 1; i++) {
    ans = (ans + dp[i][k]) % MOD;
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  calcfacs();
  while (t--) solve();
  return 0;
}