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;
template <class T>
inline T MAX(T a, T b) {
  if (a > b) return a;
  return b;
}
template <class T>
inline T MIN(T a, T b) {
  if (a < b) return a;
  return b;
}
template <class T>
inline T ABS(T x) {
  if (x < 0) return -x;
  return x;
}
inline void OPEN(const string &s) {
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
const static int inf = 1000000000;
int n, k;
int dp[2005][2005];
vector<vector<int> > fac(2005);
void gen() {
  for (int(i) = (1); (i) <= (n); ++(i)) {
    fac[i].push_back(1);
    if (i == 1) continue;
    fac[i].push_back(i);
    for (int j = 2; j < sqrt(i); j++) {
      if (i % j == 0) {
        fac[i].push_back(j);
        fac[i].push_back(i / j);
      }
    }
    if ((int)sqrt(i) * (int)sqrt(i) == i) fac[i].push_back(sqrt(i));
  }
}
int doDP(int k, int sblum) {
  if (k <= 0) return 1;
  int &ret = dp[k][sblum];
  if (ret != -1) return ret;
  ret = 0;
  for (int i = 0; i < fac[sblum].size(); i++) {
    ret = (ret + doDP(k - 1, fac[sblum][i])) % 1000000007;
  }
  return ret;
}
int main() {
  memset(dp, -1, sizeof(dp));
  scanf("%d%d", &n, &k);
  gen();
  int res = 0;
  for (int(i) = (1); (i) <= (n); ++(i))
    res = (res + doDP(k - 1, i)) % 1000000007;
  printf("%d\n", res);
  return 0;
}