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 = (long long)1e9 + 7;
vector<int> v[2020];
long long dp[2020][2020];
long long add(long long a, long long b) {
  long long c = a + b;
  while (c >= mod) c -= mod;
  return c;
}
int main() {
  long long n, m;
  scanf("%lld %lld", &n, &m);
  long long sum = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= sqrt(i); ++j) {
      if (i % j == 0) {
        v[i].push_back(j);
        if (j * j != i) {
          v[i].push_back(i / j);
        }
      }
    }
    dp[1][i] = 1;
  }
  for (int i = 2; i <= m; ++i) {
    for (int j = 1; j <= n; ++j) {
      for (int k = 0; k < v[j].size(); ++k) {
        dp[i][j] = add(dp[i][j], dp[i - 1][v[j][k]]);
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    sum = add(dp[m][i], sum);
  }
  printf("%lld\n", sum);
}