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;
long double EPS = 1e-9;
long long cel(long long a, long long b) { return ((a - 1) / b + 1); }
long long gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  return (b == 0) ? a : gcd(b, a % b);
}
long long MIN(long long a, int b) {
  long long ans;
  (a >= b) ? ans = b : ans = a;
  return ans;
}
long long MAX(long long a, int b) {
  long long ans;
  (a >= b) ? ans = a : ans = b;
  return ans;
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long po(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1) {
      ans = (ans * x) % 1000000007;
    }
    y >>= 1;
    x = (x * x) % 1000000007;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, t, x, y, m, k;
  cin >> n >> k;
  int dp[k + 1][n + 1];
  for (int i = 0; i < k + 1; i++)
    for (int j = 0; j < n + 1; j++) dp[i][j] = 0;
  for (int i = 1; i < n + 1; i++) {
    dp[1][i] = 1;
  }
  for (int i = 2; i < k + 1; i++) {
    for (int j = 1; j < n + 1; j++) {
      for (int k = j; k <= n; k += j) {
        dp[i][k] = (dp[i][k] + dp[i - 1][j]) % 1000000007;
      }
    }
  }
  x = 0;
  for (int i = 1; i < n + 1; i++) {
    x = (x + dp[k][i]) % 1000000007;
  }
  cout << x;
  return 0;
}