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 bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T1, class T2>
ostream& operator<<(ostream& s, pair<T1, T2> P) {
  return s << '<' << P.first << ", " << P.second << '>';
}
template <class T>
ostream& operator<<(ostream& s, vector<T> P) {
  for (int i = 0; i < P.size(); ++i) {
    if (i > 0) {
      s << " ";
    }
    s << P[i];
  }
  return s << endl;
}
template <class T1, class T2>
ostream& operator<<(ostream& s, map<T1, T2> P) {
  for (__typeof__((P).begin()) it = (P).begin(); it != (P).end(); ++it) {
    s << "<" << it->first << "->" << it->second << "> ";
  }
  return s << endl;
}
const int MOD = 1000000007;
void add(int& a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
int n, k;
int dp[2100][2100];
int main() {
  while (cin >> n >> k) {
    memset(dp, 0, sizeof(dp));
    dp[0][1] = 1;
    for (int i = 0; i <= k; ++i) {
      for (int j = 1; j <= n; ++j) {
        for (int l = j; l <= n; l += j) {
          add(dp[i + 1][l], dp[i][j]);
        }
      }
    }
    int res = 0;
    for (int i = 1; i <= n; ++i) add(res, dp[k][i]);
    cout << res << endl;
  }
  return 0;
}