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 dp[2010][2010];
const int mx = 2010;
int primes[1000];
int arr[2010];
const int mod = 1e9 + 7;
int maxpr;
void prep() {
  int i, j, k, l, m, n;
  for (i = 2; i < mx; i++) {
    if (!arr[i]) {
      for (j = i * i; j < mx; j += i) {
        arr[j] = i;
      }
    }
  }
  maxpr = 0, i = 1;
  while (i < mx - 5) {
    if (arr[i] == 0) {
      primes[maxpr++] = i;
    }
    i++;
  }
}
int main() {
  int i, j, k, l, m, n, t;
  prep();
  scanf("%d", &n), scanf("%d", &k);
  for (i = 1; i < 2010; i++) dp[1][i] = 1;
  for (i = 1; i < k; i++) {
    for (j = 1; j <= n; j++) {
      for (l = j; l <= n; l += j) {
        dp[i + 1][l] = (dp[i][j] + dp[i + 1][l]) % mod;
      }
    }
  }
  int ans = 0;
  for (i = 1; i <= n; i++) {
    ans = (ans + dp[k][i]) % mod;
  }
  cout << ans;
  return 0;
}