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;
template <class T>
using vec = vector<T>;
const long long inf = 1LL << 62;
const long long mod = 1e9 + 7;
const long long dx[8] = {1, 0, -1, 0, -1, 1, -1, 1};
const long long dy[8] = {0, 1, 0, -1, -1, -1, 1, 1};
const string alphabet = "abcdefghijklmnopqrstuvwxyz";
long long input() {
  long long tmp;
  cin >> tmp;
  return tmp;
}
string raw_input() {
  string tmp;
  cin >> tmp;
  return tmp;
}
template <class T>
void printx(T n) {
  cout << n;
}
template <class T, class U>
void printx(pair<T, U> p) {
  cout << "(" << p.first << "," << p.second << ")";
}
template <class T, class U, class V>
void printx(tuple<T, U, V> t) {
  cout << "{" << get<0>(t) << "," << get<1>(t) << "," << get<2>(t) << "}"
       << endl;
}
template <class T>
void printx(vector<T> v) {
  cout << "{";
  for (long long i = 0; i < v.size(); i++) {
    printx(v[i]);
    if (i != v.size() - 1) printx(",");
  }
  cout << "}";
}
template <class T>
void print(T n) {
  printx(n);
  cout << endl;
}
template <class T>
void print(set<T> s) {
  cout << "{";
  for (auto(e) = s.begin(); (e) != s.end(); (e)++) {
    if (e != s.begin()) printx(",");
    printx(*e);
  }
  cout << "}" << endl;
}
template <class T, class U>
void print(map<T, U> mp) {
  cout << "{";
  for (auto(e) = mp.begin(); (e) != mp.end(); (e)++) {
    cout << "(" << e->first << "," << e->second << ")";
  }
  cout << "}" << endl;
}
template <class T>
void printans(vec<T> v) {
  for (long long i = 0; i < (long long)v.size(); i++) {
    cout << v[i] << (i == (long long)v.size() - 1 ? "" : " ");
  }
  cout << endl;
}
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
T cut(T &a, long long l, long long r) {
  return T(a.begin() + l, a.begin() + r);
}
long long n, k;
void in() { cin >> n >> k; }
long long dp[2010][2010];
long long dfs(long long i, long long prev) {
  if (i == k) {
    return prev <= n;
  }
  if (dp[i][prev] > -1) return dp[i][prev];
  long long ans = 0;
  for (long long j = 1; prev * j <= n; j++) {
    ans += dfs(i + 1, prev * j);
    ans %= mod;
  }
  return dp[i][prev] = ans;
}
void solve() {
  memset(dp, -1, sizeof(dp));
  long long ans = 0;
  for (long long i = 1; i < n + 1; i++) {
    ans += dfs(1, i);
    ans %= mod;
  }
  print(ans);
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  in();
  solve();
}