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;
inline void read(int &x) {
  register char c = getchar();
  x = 0;
  bool neg = 0;
  for (; ((c < '0' || c > '9') && c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c >= '0' && c <= '9'; c = getchar())
    x = (x << 1) + (x << 3) + (c - '0');
  if (neg) x = -x;
}
const int MOD = 1e9 + 7;
const int MAXN = 2e3 + 5;
const long long LNF = 1e18;
const int INF = 1e9 + 10;
const int BASE = 31;
vector<int> f[MAXN];
long long dp[MAXN][MAXN];
int n, k;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  long long ans = 0;
  for (int i = 1; i <= n; ++i) dp[1][i] = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = i; j <= n; j += i) f[j].push_back(i);
  }
  for (int i = 2; i <= k; ++i) {
    for (int j = 1; j <= n; ++j) {
      for (int k = 0; k < f[j].size(); ++k) {
        dp[i][j] = dp[i][j] + dp[i - 1][f[j][k]];
        dp[i][j] %= MOD;
      }
    }
  }
  for (int i = 1; i <= n; ++i) ans = (ans + dp[k][i]) % MOD;
  cout << ans << endl;
  return 0;
}