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 = 1e9 + 7;
long long dp[3000][2000];
long long rec(long long n, long long m, long long i) {
  if (m == 0) {
    return 1;
  }
  if (dp[m][i] != -1) {
    return dp[m][i];
  }
  long long x = i;
  long long ans = 0;
  while (x <= n) {
    ans = (ans % mod + rec(n, m - 1, x) % mod) % mod;
    x += i;
  }
  dp[m][i] = ans;
  return ans;
}
void solve() {
  memset(dp, -1, sizeof(dp));
  long long n, m;
  cin >> n >> m;
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    ans = (ans % mod + rec(n, m - 1, i) % mod) % mod;
  }
  cout << ans << endl;
}
int main() {
  solve();
  return 0;
}