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 int n, t, mod = 1e9 + 7, k, h;
long long int c = 0;
int a[10];
long long int dp[2010][2010];
long long int solve(long long int k, long long int i) {
  if (k == 1) return 1;
  dp[k][i] = 0;
  for (int j = i; j <= n; j += i) {
    if (dp[k - 1][j] != -1) {
      dp[k][i] = (dp[k][i] + dp[k - 1][j]) % mod;
    } else {
      dp[k][i] = (dp[k][i] + solve(k - 1, j)) % mod;
    }
  }
  return dp[k][i] % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> k;
  memset(dp, -1, sizeof(dp));
  if (k == 1) {
    cout << n << endl;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    dp[k][i] = 0;
    for (int j = i; j <= n; j += i) {
      if (dp[k - 1][j] != -1)
        dp[k][i] = (dp[k][i] + dp[k - 1][j]) % mod;
      else {
        dp[k][i] = (dp[k][i] + solve(k - 1, j)) % mod;
      }
    }
    c = (c + dp[k][i]) % mod;
  }
  cout << c << endl;
}