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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
int mod = 1e9 + 7;
void readFile() {}
long long int power(long long int a, long long int b) {
  long long int ans = 1;
  while (b > 0) {
    if (b & 1) ans = ans * a;
    a *= a;
    b = b >> 1;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  readFile();
  int n, k;
  cin >> n >> k;
  long long int dp[k + 1][n + 1];
  memset(dp, 0, sizeof dp);
  for (int i = 1; i < n + 1; i++) dp[1][i] = 1;
  for (int i = 1; i < k + 1; i++) {
    for (int j = 1; j < n + 1; j++) {
      for (int m = 1; m * m <= j; m++) {
        if (j % m == 0) {
          dp[i][j] += (dp[i - 1][m]) % mod;
          if (m != j / m) dp[i][j] += dp[i - 1][j / m];
        }
      }
      dp[i][j] %= mod;
    }
  }
  long long int ans = 0;
  for (int i = 1; i < n + 1; i++) {
    ans += dp[k][i];
    ans %= mod;
  }
  cout << ans << '\n';
}