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;
bool cmpManh(const std::pair<long long, long long>& l,
             const std::pair<long long, long long>& r) {
  return ((llabs(l.first) + llabs(l.second)) <
          (llabs(r.first) + llabs(r.second)));
}
int gcd(int a, int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
const int N = 2001;
const int mod = 1e9 + 7;
int dp[N][N] = {0};
int main(void) {
  int n;
  scanf("%d", &n);
  int K;
  scanf("%d", &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 (int j = i; j <= n; j += i) {
        dp[l][j] = (dp[l][j] + dp[l - 1][i]) % mod;
      }
    }
  }
  long long int ans = 0;
  for (int i = 1; i <= n; ++i) {
    ans = (ans + dp[K][i] % mod) % mod;
  }
  printf("%lld\n", ans % mod);
  return (0);
}