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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
const long long MOD = 1e+9 + 7;
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int INFi = 0x7f7f7f7f;
const long long MAXN = 1e+5 + 7;
vector<long long> adj[MAXN];
long long visit[MAXN] = {};
int dx8[] = {0, 1, 1, 1, 0, -1, -1, -1}, dy8[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dx4[] = {0, 1, 0, -1}, dy4[] = {1, 0, -1, 0};
int q = 1;
long long n, k;
long long dp[2005][2005] = {};
vector<vector<long long>> div1;
long long add(long long a, long long b) { return (a + b) % MOD; }
long long mult(long long a, long long b) { return (a * b) % MOD; }
void MAIN() {
  cin >> n >> k;
  div1.resize(n + 1);
  for (long long i = 1; i <= n; i++)
    for (long long j = i; j <= n; j += i) div1[j].push_back(i);
  for (long long i = 1; i <= k; i++) {
    for (long long j = 1; j <= n; j++) {
      if (i == 1) {
        dp[i][j] = 1;
        continue;
      }
      long long len = div1[j].size();
      for (long long h = 0; h < len; h++)
        dp[i][j] = add(dp[i][j], dp[i - 1][div1[j][h]]);
    }
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) ans = add(ans, dp[k][i]);
  cout << ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(10);
  ;
  mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
  ;
  while (q--) {
    MAIN();
  }
}