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 MOD = 1000000007;
const long long N = 1010 * 2;
long long n, m, ans;
long long mem[N][N];
vector<long long> divi[N];
long long dp(long long num, long long len) {
  if (len <= 0) return 0;
  if (len == 1) return 1;
  if (mem[num][len] != -1) return mem[num][len];
  long long k = 0;
  for (auto o : divi[num]) {
    k += (dp(o, len - 1) % MOD);
  }
  return mem[num][len] = k;
}
int main() {
  cin >> n >> m;
  memset(mem, -1, sizeof mem);
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j += i) divi[j].push_back(i);
  for (int i = 1; i <= n; i++) ans += (dp(i, m) % MOD);
  cout << ans % MOD << endl;
  return 0;
}