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 <typename T>
T abs(T a) {
  return a < 0 ? -a : a;
}
const int INF = 1e9 + 7;
const double PI = 3.1415926535897932384626433832795;
const int N = 2e3 + 4;
int n, k, dp[N][N], ans;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) dp[1][i] = 1;
  for (int i = 2; i <= k; i++)
    for (int j = 1; j <= n; j++)
      for (int c = 1; c * c <= j; c++)
        if (j % c == 0) {
          dp[i][j] = (dp[i - 1][c] + 0LL + dp[i][j]) % INF;
          if (c != j / c) dp[i][j] = (dp[i - 1][j / c] + 0LL + dp[i][j]) % INF;
        }
  for (int i = 1; i <= n; i++) ans = (ans + 0LL + dp[k][i]) % INF;
  printf("%d\n", ans);
  return 0;
}