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 MOD = int(1e9) + 7;
int DP[2001][2001];
vector<int> Div[2001];
int main() {
  for (int i = 2; i <= 2000; i++) {
    for (int j = 1; j * j <= i; j++)
      if (i % j == 0) {
        Div[i].push_back(j);
        if (j != i / j) Div[i].push_back(i / j);
      }
  }
  int n, k;
  scanf("%d", &n);
  scanf("%d", &k);
  for (int i = 1; i <= n; i++) DP[i][1] = 1;
  for (int j = 2; j <= k; j++) {
    DP[1][j] = 1;
    for (int i = 2; i <= n; i++)
      for (int k = 0; k < (int)(Div[i].size()); k++)
        DP[i][j] = (DP[i][j] + DP[Div[i][k]][j - 1]) % MOD;
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) ans = (ans + DP[i][k]) % MOD;
  printf("%d\n", ans);
  return 0;
}