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;
using pi = pair<long long, long long>;
using vi = vector<long long>;
using vvi = vector<vector<long long> >;
const long long NUM = 1000030;
const long long N = 10000000;
vector<long long> lp, sieve;
vector<long long> pr;
void calc_sieve() {
  sieve.resize(NUM + 1, 0);
  for (long long x = 2; x <= NUM; x++) {
    if (sieve[x]) continue;
    for (long long u = x; u <= NUM; u += x) {
      sieve[u] = x;
    }
  }
}
void primefactor() {
  lp.resize(N + 1, 0);
  for (long long i = 2; i <= N; ++i) {
    if (lp[i] == 0) {
      lp[i] = i;
      pr.push_back(i);
    }
    for (long long j = 0;
         j < (long long)pr.size() && pr[j] <= lp[i] && i * pr[j] <= N; ++j)
      lp[i * pr[j]] = pr[j];
  }
}
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long binpow(long long a, long long b, long long mod) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res % mod;
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return ((a / gcd(a, b)) * b); }
bool comp(long long a, long long b) { return a > b; }
long long inversemod(long long a, long long mod) {
  return binpow(a, mod - 2, mod);
}
void addEdge(long long u, long long v, vector<long long> adj[]) {
  adj[u].push_back(v);
  adj[v].push_back(u);
}
const long long mod = 1e9 + 7;
void solve() {
  long long n, K;
  cin >> n >> K;
  vvi dp(n + 1, vi(K + 1, 0));
  for (long long i = 1; i <= n; i++) {
    dp[i][1] = 1;
  }
  for (long long t = 2; t <= K; t++) {
    for (long long i = 1; i <= n; i++) {
      for (long long k = i; k <= n; k += i) {
        dp[k][t] = (dp[k][t] + dp[i][t - 1]) % mod;
      }
    }
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    ans = (ans + dp[i][K]) % mod;
  }
  if (ans < 0) {
    ans += mod;
  }
  cout << ans << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long test_case = 1;
  long long tc = 1;
  while (test_case--) {
    solve();
    tc++;
  }
  return 0;
}