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 mod = 1000000007;
vector<int> prims;
int primes[2001];
long long dp[21][2001];
long long fun(int N, int K) {
  long long ans = 1;
  for (int i = 0; i < prims.size(); i++) {
    if (N % prims[i] == 0) {
      int cont = 0;
      while (N % prims[i] == 0) {
        cont++;
        N /= prims[i];
      }
      ans = (ans * dp[cont][K]) % mod;
    }
  }
  return ans;
}
void init() {
  for (int i = 2; i < 2001; i++) {
    primes[i] = 1;
  }
  for (int i = 2; i < 2001; i++) {
    if (!primes[i]) continue;
    for (int j = i * i; j < 2001; j += i) {
      primes[j] = 0;
    }
  }
  for (int i = 2; i < 2001; i++) {
    if (primes[i]) prims.push_back(i);
  }
  for (int l = 1; l < 2001; l++) {
    for (int p = 0; p < 21; p++) {
      dp[p][l] = 0;
      if (l == 1 || p == 0)
        dp[p][l] = 1;
      else {
        for (int p2 = 0; p2 <= p; p2++) {
          dp[p][l] = (dp[p][l] + dp[p2][l - 1]) % mod;
        }
      }
    }
  }
}
int main() {
  init();
  int N, K;
  scanf("%d%d", &N, &K);
  long long ans = 0;
  for (int i = 1; i <= N; i++) {
    ans = (ans + fun(i, K)) % mod;
  }
  cout << ans << endl;
  return 0;
}