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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const double pi = acos(-1.0), eps = 1e-6;
void File() {
  freopen("D:\\in.txt", "r", stdin);
  freopen("D:\\out.txt", "w", stdout);
}
template <class T>
inline void read(T &x) {
  char c = getchar();
  x = 0;
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
}
const int maxn = 2020;
long long dp[maxn][maxn], mod = 1e9 + 7;
int n, m;
int main() {
  scanf("%d%d", &m, &n);
  for (int i = 1; i <= m; i++) dp[1][i] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      for (int k = j; k <= m; k = k + j) {
        dp[i + 1][k] = (dp[i + 1][k] + dp[i][j]) % mod;
      }
    }
  }
  long long ans = 0;
  for (int i = 1; i <= m; i++) ans = (ans + dp[n][i]) % mod;
  cout << ans << endl;
  return 0;
}