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 = int(1e5);
const int INF = int(1e9);
const int MOD = int(1e9) + 7;
long long n, k, d[2015][2015], res;
vector<long long> v[2015];
long long calc(long long x, long long len) {
  long long sum = 0;
  for (int i = 0; i < v[x].size(); ++i) {
    sum += d[v[x][i]][len - 1];
    sum %= MOD;
  }
  return sum;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    d[i][1] = 1;
  }
  for (int i = 1; i <= k; ++i) {
    d[1][i] = 1;
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j * j <= i; j++) {
      if (i % j == 0) {
        v[i].push_back(j);
        v[i].push_back(i / j);
      }
    }
    v[i].erase(unique(v[i].begin(), v[i].end()), v[i].end());
  }
  for (int i = 2; i <= n; ++i) {
    for (int j = 2; j <= k; ++j) {
      d[i][j] = calc(i, j);
    }
  }
  for (int i = 1; i <= n; ++i) {
    res += d[i][k];
    res %= MOD;
  }
  cout << res;
  return 0;
}