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 = 1e9 + 7;
const long long INF = 0x7FFFFFFFFFFFFFFF / 2;
long long n, k;
long long dp[2001][2001] = {0};
vector<int> factors[2001];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    for (int z = 1; z <= (int)sqrt(i); z++) {
      if (i % z == 0) {
        int f1 = z;
        int f2 = i / z;
        factors[i].push_back(f1);
        if (f1 != f2) factors[i].push_back(f2);
      }
    }
    for (int j = 1; j <= k; j++) {
      if (j == 1 || i == 1) dp[i][j] = 1;
    }
  }
  for (int i = 2; i <= n; i++) {
    for (int j = 2; j <= k; j++) {
      for (auto lul : factors[i]) {
        dp[i][j] = (dp[i][j] + dp[lul][j - 1]) % mod;
      }
    }
  }
  long long ans = 0;
  for (int i = n; i >= 1; i--) {
    ans = (ans + dp[i][k]) % mod;
  }
  cout << ans;
  return 0;
}