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>
T MAX(T a, T b) {
  return a > b ? a : b;
}
template <class T>
T MIN(T a, T b) {
  return a < b ? a : b;
}
template <class T1>
void __(T1 p) {
  cout << p << endl;
}
template <class T1>
void deb(T1 p) {
  cout << "Debugging: " << p << endl;
}
template <class T1, class T2>
void deb(T1 p, T2 q) {
  cout << "Debugging: " << p << "\t" << q << endl;
}
template <class T1, class T2, class T3>
void deb(T1 p, T2 q, T3 r) {
  cout << "Debugging: " << p << "\t " << q << "\t " << r << endl;
}
template <class T1, class T2, class T3, class T4>
void deb(T1 p, T2 q, T3 r, T4 s) {
  cout << "Debugging: " << p << "\t " << q << "\t " << r << "\t " << s << endl;
}
long long int POOW(long long b, long long p) {
  if (p == 0) return 1;
  return b * POOW(b, p - 1);
}
const int first[] = {0, 0, 1, -1, -1, 1, -1, 1};
const int second[] = {1, -1, 0, 0, 1, 1, -1, -1};
const int kx[] = {-2, -1, 1, 2, 2, 1, -1, -2};
const int ky[] = {1, 2, 2, 1, -1, -2, -2, -1};
vector<int> v[2003];
int n, k;
long long ans, dp[2003][2003];
long long GO(int pos, int t) {
  if (pos == 1) return 1ll;
  long long &ret = dp[pos][t];
  if (~ret) return ret;
  ret = 0;
  for (int i = v[t].size() - 1; i >= 0; i--) {
    ret += GO(pos - 1, v[t][i]);
    ret %= 1000000007;
  }
  return ret;
}
int main() {
  int i, t, j, k, l, keis(0), c, d, x, y, a, b;
  for (i = 1; i < 2001; i++) {
    b = i >> 1;
    for (j = 1; j <= b; j++) {
      if (i % j == 0) v[i].push_back(j);
    }
    v[i].push_back(i);
  }
  cin >> n >> k;
  memset(dp, -1, sizeof dp);
  ans = 0;
  for (i = 1; i <= n; i++) {
    ans += GO(k, i);
    ans %= 1000000007;
  }
  __(ans);
  return 0;
}