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;
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};
long long toInt(string s) {
  stringstream ss(s);
  long long k;
  ss >> k;
  return k;
}
string toString(long long k) {
  stringstream ss;
  ss << k;
  return ss.str();
}
bool prime(long long x) {
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) return false;
  }
  return true;
}
long long n, m, k;
long long dp[2222][2222];
int main(void) {
  cin >> n >> k;
  for (int i = (1); i <= (n); i++) dp[1][i] = 1;
  for (int l = (2); l <= (k); l++) {
    for (int i = (1); i <= (n); i++) {
      for (long long j = i; j <= n; j += i) {
        dp[l][i] += dp[l - 1][j];
        dp[l][i] %= 1000000007;
      }
    }
  }
  long long ans = 0;
  for (int i = (1); i <= (n); i++) {
    ans += dp[k][i];
    ans %= 1000000007;
  }
  cout << ans << endl;
  return 0;
}