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;
long long mod = 1000000000 + 7;
vector<vector<long long> > divi;
void sieve() {
  divi.assign(2010, vector<long long>());
  for (int i = 1; i < 2010; ++i) {
    for (int j = i; j < 2010; j += i) {
      divi[i].push_back(j);
    }
  }
}
int main() {
  sieve();
  int n, k;
  cin >> n >> k;
  long long dp[k + 1][n + 1];
  for (int i = 0; i <= k; ++i) {
    for (int j = 0; j <= n; ++j) {
      dp[i][j] = 0;
    }
  }
  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) {
      for (int l = 0; l < divi[j].size(); l++) {
        if (divi[j][l] <= n)
          dp[i][j] = (dp[i][j] + dp[i - 1][divi[j][l]]) % mod;
        else
          break;
      }
    }
  }
  long long res = 0;
  for (int i = 1; i <= n; ++i) {
    res = (res + dp[k][i]) % mod;
  }
  cout << res << endl;
  return 0;
}