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;
#pragma comment(linker, "/STACK:64000000")
const int maxn = 2020;
const int inf = 1000000007;
const int mod = 1000000007;
const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, 1, -1};
vector<int> e[maxn];
int dp[maxn][maxn];
void precalc() {
  for (int i = 1; i <= 2000; i++) {
    for (int j = i; j <= 2000; j += i) {
      e[j].push_back(i);
    }
  }
}
int main() {
  precalc();
  int n, k;
  while (cin >> n >> k) {
    for (int i = 0; i <= n; i++)
      for (int j = 0; j <= k; j++) dp[j][i] = 0;
    dp[0][1] = 1;
    for (int i = 1; i <= k; i++) {
      for (int j = 1; j <= n; j++) {
        for (int l = 0; l < e[j].size(); l++) {
          dp[i][j] = dp[i][j] + dp[i - 1][e[j][l]];
          if (dp[i][j] >= mod) dp[i][j] -= mod;
        }
      }
    }
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      ans += dp[k][i];
      if (ans >= mod) ans -= mod;
    }
    cout << ans << endl;
  }
  return 0;
}