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;
long long MINI = -1000000000000000000;
long long MAXI = 1000000000000000000;
long long dp[2005][2005];
vector<vector<long long>> divs;
long long n, k;
void prepare() {
  long long j;
  for (long long i = 1; i <= n; i++) {
    vector<long long> v;
    for (j = 1; j * j <= i; j++) {
      if (i % j == 0) {
        if (i / j == j) {
          v.push_back(j);
        } else {
          v.push_back(j);
          v.push_back(i / j);
        }
      }
    }
    divs.push_back(v);
  }
}
int main() {
  cin >> n >> k;
  prepare();
  for (int i = 0; i <= n; i++) {
    dp[1][i] = 1;
  }
  for (int i = 2; i <= k; i++) {
    for (int j = 1; j <= n; j++) {
      for (int k = 0; k < divs[j - 1].size(); k++) {
        dp[i][j] += dp[i - 1][divs[j - 1][k]];
        dp[i][j] %= 1000000007;
      }
    }
  }
  long long sum = 0;
  for (int i = 1; i <= n; i++) {
    sum += dp[k][i];
    sum %= 1000000007;
  }
  cout << sum;
}