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 N = 1e6 + 2;
const long long MOD = 1e9 + 7;
const long long inf = 1e18 + 6;
const long long M = 2003;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
long long dp[M][M];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(20);
  cout << fixed;
  long long n, x;
  cin >> n >> x;
  dp[0][1] = 1;
  for (int i = (int)(1); i <= (int)(x); ++i)
    for (int j = (int)(1); j <= (int)(n); ++j)
      for (int k = j; k <= n + 3; k += j)
        dp[i][k] = (dp[i][k] + dp[i - 1][j]) % MOD;
  long long ans = 0;
  for (int i = (int)(1); i <= (int)(n); ++i) ans = (ans + dp[x][i]) % MOD;
  cout << ans;
}