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 = 1000000007;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1);
const double EPS = 1e-10;
const int N = 2010;
int n, k;
vector<int> divisors[N];
long long dp[N][N];
long long getDp(int i, int j) {
  long long &res = dp[i][j];
  if (res != -1) {
    return res;
  }
  if (j == 1) {
    return res = 1;
  }
  res = 0;
  for (int t : divisors[i]) {
    res = (res + getDp(t, j - 1)) % MOD;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= (int)floor(sqrt(i)); ++j) {
      if (i % j == 0) {
        divisors[i].push_back(j);
        if (j != i / j) {
          divisors[i].push_back(i / j);
        }
      }
    }
  }
  memset(dp, -1, sizeof(dp));
  long long res = 0;
  for (int i = 1; i <= n; ++i) {
    res = (res + getDp(i, k)) % MOD;
  }
  cout << res;
}