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;
const int MAX = 123456789;
const double EPS = 1e-10;
const double PI = 2 * asin(1.0);
const long long mod = 1e9 + 7;
inline int cmp(double x, double y = 0, double tol = EPS) {
  return (x <= y + tol) ? (x + tol < y) ? -1 : 0 : 1;
}
long long pd[2001][2001];
vector<int> divi[2001];
void sieve() {
  for (int i = 1; i < 2001; ++i) {
    for (int j = i; j >= 1; --j) {
      if (i % j == 0) divi[i].push_back(j);
    }
  }
}
int main() {
  sieve();
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n + 1; ++i) pd[1][i] = 1;
  for (int i = 2; i < k + 1; ++i) {
    for (int j = 1; j < n + 1; ++j) {
      for (int l = 0; l < ((int)divi[j].size()); ++l) {
        pd[i][j] += pd[i - 1][divi[j][l]];
        pd[i][j] %= mod;
      }
    }
  }
  long long sum = 0;
  for (int i = 1; i < n + 1; ++i) {
    sum += pd[k][i];
    sum %= mod;
  }
  cout << sum << endl;
}