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 mod = 1000000007;
const int Max = 2002;
long long dp[Max][Max];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < k; i++) {
    for (int j = 0; j <= n; j++) dp[i][j] = 0;
  }
  for (int i = 1; i <= n; i++) dp[0][i] = 1;
  for (int i = 1; i < k; i++) {
    for (int j = 1; j <= n; j++) {
      if (j == 1) {
        (dp[i][j] += dp[i - 1][j]) %= mod;
      } else {
        for (int d = 1; d * d <= j; d++) {
          if (j % d == 0) {
            if (d != j / d) {
              dp[i][j] =
                  ((dp[i][j] + dp[i - 1][d]) % mod + dp[i - 1][j / d]) % mod;
            } else
              (dp[i][j] += dp[i - 1][d]) %= mod;
          }
        }
      }
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) (ans += dp[k - 1][i]) %= mod;
  cout << ans << endl;
}