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 inf = 0x3f3f3f3f3f3f3f3f;
long long power(long long a, long long b) {
  long long res = 1;
  while (b != 0) {
    if (b & 1) {
      res *= a;
    }
    a = (a * a);
    b = (b / 2);
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t;
  t = 1;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long dp[n + 1][k + 1];
    memset(dp, (long long)0, sizeof(dp));
    for (long long i = 1; i <= n; i++) dp[i][1] = 1;
    long long total = 0;
    for (long long i = 1; i <= k; i++) {
      for (long long j = 1; j <= n; j++) {
        if (dp[j][i] != 0) {
          for (long long m = j; m <= n; m += j)
            dp[m][i + 1] = (dp[m][i + 1] + dp[j][i]) % 1000000007;
        }
      }
    }
    for (long long i = 1; i <= n; i++) {
      total = (total + dp[i][k]) % 1000000007;
    }
    cout << total;
  }
  return 0;
}