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 max(long long a, long long b) {
  if (a > b) {
    return a;
  } else {
    return b;
  }
}
long long min(long long a, long long b) {
  if (a < b) {
    return a;
  } else {
    return b;
  }
}
long long power(long long b, long long e) {
  if (e == 0) return 1;
  if (e % 2)
    return ((b * power((b) * (b), (e - 1) / 2)));
  else
    return power((b) * (b), e / 2);
}
long long modpower(long long b, long long e) {
  long long MOD = 2e18;
  if (e == 0) return 1;
  if (e % 2)
    return ((b % MOD) * power((b % MOD) * (b % MOD), (e - 1) / 2)) % MOD;
  else
    return power((b % MOD) * (b % MOD), e / 2) % MOD;
}
void dpv(vector<long long> v) {
  for (long long i = 0; i < v.size(); i++) {
    cout << v[i] << " ";
  }
  cout << '\n';
}
void dpv(vector<pair<long long, long long> > v) {
  for (long long i = 0; i < v.size(); i++) {
    cout << v[i].first << " " << v[i].second << '\n';
  }
}
void dpv(set<long long> v) {
  for (auto i : v) {
    cout << i << " ";
  }
  cout << '\n';
}
const long long MOD = 1e9 + 7;
const long long ms = 2020;
long long dp[ms][ms];
void oblivious() {
  long long n, k;
  cin >> k >> n;
  for (long long i = 1; i <= k; i++) {
    dp[1][i] = 1;
  }
  for (long long i = 2; i <= n; i++) {
    for (long long j = 1; j <= k; j++) {
      for (long long kk = j; kk <= k; kk += j) {
        dp[i][kk] =
            (dp[i][kk] + dp[i - 1][j] >= MOD ? dp[i][kk] + dp[i - 1][j] - MOD
                                             : dp[i][kk] + dp[i - 1][j]);
      }
    }
  }
  long long ans = 0;
  for (long long i = 1; i <= k; i++) {
    ans = (ans + dp[n][i] >= MOD ? ans + dp[n][i] - MOD : ans + dp[n][i]);
  }
  std::cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    oblivious();
  }
  return 0;
}