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;
void debug(vector<long long int> v, int a) {
  cout << "DEBUG " << a << endl;
  for (auto k = v.begin(); k != v.end(); k++) cout << *k << " ";
  cout << endl;
}
long long int dp[2002][2002];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t, n, m, temp, temp1, temp2, temp3, k, a, b, c, d, count, x, y;
  long long int mod = 1e9 + 7;
  cin >> n >> k;
  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++) {
      long long int sq = sqrt(j);
      for (int m = 1; m <= sq; m++) {
        if (j % m == 0) {
          dp[i][j] = ((dp[i][j] % mod) + (dp[i - 1][m] % mod)) % mod;
          if ((j / m) != m) {
            dp[i][j] = ((dp[i][j] % mod) + (dp[i - 1][j / m] % mod)) % mod;
          }
        }
      }
    }
  }
  long long int sum = 0;
  for (int i = 1; i <= n; i++) {
    sum = ((sum % mod) + (dp[k][i] % mod)) % mod;
  }
  cout << sum;
}