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;
int D[2005][2005], N, K;
void precalc() {
  for (int i = 1; i <= N; i++) D[i][1] = 1;
  for (int l = 1; l <= K; l++) {
    for (int x = 0; x <= N; x++) {
      for (int divideby = 1; divideby * divideby <= x; divideby++) {
        if (x % divideby == 0) {
          D[x][l] += D[x / divideby][l - 1];
          D[x][l] %= 1000000007;
          if (divideby * divideby != x) {
            D[x][l] += D[divideby][l - 1];
            D[x][l] %= 1000000007;
          }
        }
      }
    }
  }
}
int main() {
  scanf("%d %d", &N, &K);
  precalc();
  int sol = 0;
  for (int i = 0; i <= N; i++) {
    sol = sol + D[i][K];
    sol %= 1000000007;
  }
  printf("%d\n", sol);
  return 0;
}