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 mpow(long long a, long long n) {
  long long ret = 1;
  long long b = a;
  while (n) {
    if (n & 1) ret = (ret * b) % 1000000007;
    b = (b * b) % 1000000007;
    n >>= 1;
  }
  return (long long)ret;
}
long long dp[5009][5009] = {};
vector<int> fac[5009];
long long rec(long long n, long long k) {
  if (dp[n][k] != -1) return dp[n][k];
  if (k == 0) return 0;
  if (n == 1 || k == 1) return dp[n][k] = 1;
  long long ans = 0;
  int i, p = fac[n].size();
  for (i = 0; i < p; i++) {
    ans += (rec(fac[n][i], k - 1));
    if (ans >= 1000000007) ans -= 1000000007;
  }
  return dp[n][k] = ans;
}
void gen() {
  int j;
  for (int i = 2; i <= 5000; i++) {
    for (j = 1; j * j < i; j++)
      if (i % j == 0) fac[i].push_back(j), fac[i].push_back(i / j);
    if (j * j == i) fac[i].push_back(j);
  }
}
int main() {
  gen();
  memset(dp, -1, sizeof(dp));
  long long n, k;
  cin >> n >> k;
  long long ans = 0;
  for (int i = n; i >= 1; i--) {
    ans += rec(i, k);
    if (ans >= 1000000007) ans -= 1000000007;
  }
  cout << ans << endl;
  return 0;
}