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 int pow(long long int x, long long int y, long long int m) {
  long long int res = 1;
  x = x % m;
  while (y > 0) {
    if (y & 1) res = ((res % m) * (x % m)) % m;
    y = y >> 1;
    x = ((x % m) * (x % m)) % m;
  }
  return res % m;
}
template <typename Arg1>
void ZZ(const char* name, Arg1&& arg1) {
  std::cerr << name << " = " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void ZZ(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " = " << arg1;
  ZZ(comma, args...);
}
const long long int INF = 0xFFFFFFFFFFFFFFFL;
clock_t time_p = clock();
void abhigyan10() {
  time_p = clock() - time_p;
  cerr << "Time Taken : " << (float)(time_p) / CLOCKS_PER_SEC << "\n";
}
long long int seed;
mt19937 rng(seed = chrono::steady_clock::now().time_since_epoch().count());
inline long long int rnd(long long int l = 0, long long int r = INF) {
  return uniform_int_distribution<long long int>(l, r)(rng);
}
const long long int mod = 1000000007;
int dp[2002][2002];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, k, i, j, ans = 0, l, x;
  cin >> n >> k;
  for (i = 1; i <= n; i++) dp[1][i] = 1;
  for (i = 2; i <= k; i++) {
    for (j = 1; j <= n; j++) {
      for (l = j; l <= n; l += j)
        dp[i][l] = (dp[i - 1][j] % mod + dp[i][l] % mod) % mod;
    }
  }
  for (i = 1; i <= n; i++) {
    ans = (ans % mod + dp[k][i] % mod) % mod;
  }
  cout << ans;
  return 0;
}