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;
inline int read() {
  int x = 0;
  char ch = getchar();
  bool positive = 1;
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') positive = 0;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  return positive ? x : -x;
}
char RC() {
  char c = getchar();
  while (c == ' ' || c == '\n') c = getchar();
  return c;
}
const int mx = 2005;
const int inf = 2100000000;
const int mod = 1000000007;
int gcd(int x, int y) { return (y == 0) ? x : gcd(y, x % y); }
int n, m;
vector<int> v[mx];
int dp[mx][mx];
void pre() {
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= i; ++j) {
      if (i % j == 0) v[i].push_back(j);
    }
  }
}
int main() {
  cin >> n >> m;
  pre();
  for (int i = 1; i <= m; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (i == 1) {
        dp[i][j] = 1;
        continue;
      }
      for (int k = 0; k < v[j].size(); ++k) {
        dp[i][j] = (dp[i][j] + dp[i - 1][v[j][k]]) % mod;
      }
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    ans = (ans + dp[m][i]) % mod;
  }
  cout << ans << endl;
  return 0;
}