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;
double pi = 3.1415926536;
const int oo = (int)1e9;
const long long OO = numeric_limits<long long>::max();
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
int di[] = {0, 0, 1, -1, 1, -1, 1, -1};
int dj[] = {1, -1, 0, 0, 1, -1, -1, 1};
int f1[] = {0, 0, 0, 1, 1, 1, -1, -1, -1};
int f2[] = {-1, 0, 1, -1, 0, 1, -1, 0, 1};
int dp[2001][2001];
int n, k;
int solve(int x, int k) {
  if (!k) {
    return 1;
  }
  int &rem = dp[x][k];
  if (rem) {
    return rem;
  }
  for (int i = x; i <= n; i += x) {
    rem = (rem + solve(i, k - 1)) % (oo + 7);
  }
  return rem;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  cout << solve(1, k);
  return 0;
}