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 f[2005][2005];
const int mod = (int)1e9 + 7;
int main() {
  int N, K;
  cin >> N >> K;
  long long sum = 0;
  for (int l = 1; l <= K; l++) {
    for (int d = 1; d <= N; d++) {
      if (l == 1)
        f[l][d] = 1;
      else {
        for (int i = 1; i <= sqrt(d); i++)
          if (d % i == 0) {
            f[l][d] += f[l - 1][i];
            f[l][d] %= mod;
            if (d != (i * i)) {
              f[l][d] += f[l - 1][d / i];
              f[l][d] %= mod;
            }
          }
      }
      if (l == K) sum = (sum + f[l][d]) % mod;
    }
  }
  cout << sum << endl;
}