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 long dp[2001][2001], maxn;
vector<long long> factor[2005];
long long ways(long long n, long long k) {
  if (k == 1) {
    dp[1][n] = 1;
    return 1;
  }
  if (n < 1) return 0;
  if (dp[k][n] != -1) return dp[k][n];
  long long ans = 0;
  for (auto j : factor[n]) {
    ans += ways(j, k - 1);
    ans %= 1000000007;
  }
  dp[k][n] = ans;
  return ans;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n, k;
  cin >> n >> k;
  for (long long i = 0; i < k + 1; i++) {
    for (long long j = 0; j < n + 1; j++) dp[i][j] = -1;
  }
  maxn = n;
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    for (long long j = i; j <= n; j += i) {
      factor[j].push_back(i);
    }
  }
  for (long long i = 1; i <= n; i++) {
    long long c = ways(i, k);
    ans += c;
    ans %= 1000000007;
  }
  cout << ans << endl;
  return 0;
}