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 int prime[2001];
vector<long long int> factor[2001];
void seive(long long int n) {
  memset(prime, 0, sizeof(prime));
  prime[0] = 0;
  prime[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (prime[i] == 0) {
      prime[i] = i;
      for (int j = 2 * i; j <= n; j += i) {
        if (prime[j] == 0) prime[j] = i;
      }
    }
  }
}
void factors(long long int n) {
  for (long long int i = 2; i <= n; i++) {
    for (long long int j = 1; j * j <= i; j++) {
      if (i % j == 0) {
        factor[i].push_back(j);
        if (i / j != j) factor[i].push_back(i / j);
      }
    }
  }
}
long long int add(long long int a, long long int b) {
  return ((a % 1000000007) + (b % 1000000007)) % 1000000007;
}
long long int subt(long long int a, long long int b) {
  if (a % 1000000007 - b % 1000000007 >= 0)
    return ((a % 1000000007) - (b % 1000000007)) % 1000000007;
  else
    return ((a % 1000000007) - (b % 1000000007) + 1000000007) % 1000000007;
}
long long int mul(long long int a, long long int b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int n, k;
  cin >> n >> k;
  long long int dp[n + 1][k + 1];
  memset(dp, 0, sizeof(dp));
  for (long long int i = 1; i <= n; i++) dp[i][1] = 1;
  for (long long int i = 1; i <= k; i++) dp[1][i] = 1;
  factors(2000);
  factor[1].push_back(1);
  for (long long int i = 2; i <= n; i++) {
    for (long long int j = 2; j <= k; j++) {
      long long int t = 0;
      for (long long int l = 0; l < factor[i].size(); l++)
        t = add(t, dp[factor[i][l]][j - 1]);
      dp[i][j] = t % 1000000007;
    }
  }
  long long int ans = 0;
  for (long long int i = 1; i <= n; i++) ans = add(ans, dp[i][k]);
  cout << ans;
}