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 c[2005];
int dp[2005][2005];
void sieve() {
  c[1] = 1;
  for (int i = 2; i <= sqrt(2005); i++) {
    if (c[i] == 0) {
      c[i] = i;
      for (int j = i * 2; j <= 2005; j += i) {
        if (c[j] == 0) {
          c[j] = i;
        }
      }
    }
  }
}
int comb(int val, int k) {
  if (k == 1) {
    return 1;
  }
  if (dp[val][k] != -1) {
    return dp[val][k] % 1000000007;
  }
  long long cc = 0;
  long long h = val;
  cc = (cc % 1000000007 + comb(val, k - 1) % 1000000007) % 1000000007;
  while (val != 1) {
    val = val / c[val];
    cc = (cc % 1000000007 + comb(val, k - 1) % 1000000007) % 1000000007;
  }
  return dp[h][k] = cc;
}
int main() {
  memset(c, 0, sizeof(c));
  sieve();
  memset(dp, 0, sizeof(dp));
  long long n, k;
  cin >> n >> k;
  long long count = 0;
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= n; j++) {
      if (i == 1) {
        dp[i][j] = 1;
      } else {
        for (int k = j; k <= n; k += j) {
          dp[i][k] =
              (dp[i][k] % 1000000007 + dp[i - 1][j] % 1000000007) % 1000000007;
        }
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    count = (count % 1000000007 + dp[k][i] % 1000000007) % 1000000007;
  }
  cout << count << endl;
}