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 int maxn = 2007, mod = 1e9 + 7;
int n, k, a[maxn][maxn], q[maxn], ans = 0;
void solve() {
  cin >> n >> k;
  for (int i = 1; i < maxn; ++i) {
    a[i][1] = 1;
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1, curr, qs; j <= k; ++j) {
      curr = sqrt(i), qs = 0;
      for (int l = 1; l <= curr; ++l) {
        if (i % l == 0) {
          q[qs++] = l;
          a[i][j] = (a[l][j - 1] + a[i][j]) % mod;
        }
      }
      for (int l = 0; l < qs; ++l) {
        if (q[l] * q[l] != i) {
          a[i][j] = (a[i / q[l]][j - 1] + a[i][j]) % mod;
        }
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    ans += a[i][k], ans %= mod;
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}