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 min(long long a, long long b) {
  if (a < b)
    return a;
  else
    return b;
}
long long max(long long a, long long b) {
  if (a < b)
    return b;
  else
    return a;
}
long long min3(long long a, long long b, long long c) {
  return min(a, min(b, c));
}
long long max3(long long a, long long b, long long c) {
  return max(a, max(b, c));
}
long long ceil1(long long x, long long y) {
  if (x % y)
    return (x + y) / y;
  else
    return x / y;
}
long long NC = 100;
long long N = 1e9 + 7;
const long long S = 3e5 + 7;
long long mod(long long n);
long long gcd(long long a, long long b);
long long modM(long long n, long long m);
long long modA(long long n, long long m);
long long modS(long long n, long long m);
long long power(long long a, long long b);
void pra(long long a[], long long n);
void ipgraph(long long n, long long m);
vector<vector<long long> > v;
vector<bool> vis;
long long dp[2005][2005];
void solve() {
  memset(dp, 0, sizeof(dp));
  long long n, k;
  cin >> n >> k;
  for (long long i = 1; i < n + 1; i++) dp[1][i] = 1;
  for (long long i = 1; i < k + 1; i++) {
    for (long long j = 1; j < n + 1; j++) {
      for (long long l = j; l <= n; l += j)
        dp[i + 1][l] += dp[i][j], dp[i + 1][l] %= N;
    }
  }
  long long ans = 0;
  for (long long i = 1; i < n + 1; i++) ans += dp[k][i], ans %= N;
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << setprecision(15);
  long long t = 1;
  while (t--) solve();
  cerr << '\n'
       << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
  return 0;
}
void ipgraph(long long n, long long m) {
  v.assign(n, {});
  vis.assign(n, 0);
  long long x, y;
  for (long long i = 0; i < m; i++) {
    cin >> x >> y;
    x--, y--;
    v[y].push_back(x);
    v[x].push_back(y);
  }
}
long long gcd(long long a, long long b) {
  if (b > a) return gcd(b, a);
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long power(long long a, long long b) {
  if (b == 0) return 1;
  long long c = power(a, b / 2);
  if (b % 2 == 0)
    return modM(c, c);
  else
    return modM(modM(c, c), a);
}
void pra(long long a[], long long n) {
  for (long long i = 0; i < n; i++) cout << a[i] << " ";
  cout << '\n';
}
long long mod(long long n) { return (n % N + N) % N; }
long long modM(long long n, long long m) { return ((n % N * m % N) + N) % N; }
long long modA(long long n, long long m) { return ((n % N + m % N) + N) % N; }
long long modS(long long n, long long m) { return ((n % N - m % N) + N) % N; }