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 namespace std::chrono;
inline long long int max(long long int a, long long int b) {
  return a > b ? a : b;
}
inline long long int min(long long int a, long long int b) {
  return a < b ? a : b;
}
inline long long int max(vector<long long int> a) {
  return *max_element(a.begin(), a.end());
}
inline long long int min(vector<long long int> a) {
  return *min_element(a.begin(), a.end());
}
inline long long int min(long long int a[], long long int n) {
  return *min_element(a, a + n);
}
inline long long int max(long long int a[], long long int n) {
  return *max_element(a, a + n);
}
inline long long int sum(long long int a[], long long int n) {
  long long int s = 0;
  for (long long int i = 0; i < (n); ++i) s += a[i];
  return s;
}
inline long long int sum(vector<long long int> a) {
  long long int s = 0;
  for (long long int i = 0; i < (a.size()); ++i) s += a[i];
  return s;
}
const long long inf = 1e18;
double pie = 2 * acos(0.0);
void show(vector<long long int> v) {
  for (long long int i = 0; i < (v.size()); ++i) cout << v[i] << " ";
  cout << "\n";
}
void show(vector<int> v) {
  for (long long int i = 0; i < (v.size()); ++i) cout << v[i] << " ";
  cout << "\n";
}
void show(int a[], int arraysize) {
  for (int i = 0; i < arraysize; ++i) cout << a[i] << " ";
  cout << "\n";
}
void show(long long int a[], long long int arraysize) {
  for (long long int i = 0; i < arraysize; ++i) cout << a[i] << " ";
  cout << "\n";
}
void show(bool a[], long long int arraysize) {
  for (long long int i = 0; i < arraysize; ++i) cout << a[i] << " ";
  cout << "\n";
}
void show(set<int> s) {
  for (auto it = s.begin(); it != s.end(); it++) {
    cout << *it << " ";
  }
  cout << "\n";
}
void show(map<long long int, long long int> mp) {
  for (auto it = mp.begin(); it != mp.end(); it++) {
    cout << it->first << " " << it->second << "\n";
  }
  cout << "\n";
}
bool comp(pair<long long int, long long int> p1,
          pair<long long int, long long int> p2) {
  return p1.first > p2.first;
}
long long int modpow(long long int x, long long int n) {
  long long int res = 1;
  while (n > 0) {
    if (n & 1) res = ((res % 1000000007) * (x % 1000000007)) % 1000000007;
    x = ((x % 1000000007) * (x % 1000000007)) % 1000000007;
    n >>= 1;
  }
  return res;
}
long long int power(long long int x, long long int n) {
  long long int res = 1;
  while (n > 0) {
    if (n & 1) res = res * x;
    x = x * x;
    n >>= 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(12);
  long long int n, k;
  cin >> n >> k;
  vector<long long int> a[n + 1];
  a[1].push_back(1);
  for (long long int i = 2; i < (n + 1); ++i) {
    a[i].push_back(1);
    a[i].push_back(i);
    for (long long int j = 2; j <= sqrt(i); ++j) {
      if (i % j == 0) {
        a[i].push_back(j);
        if (i / j != j) a[i].push_back(i / j);
      }
    }
  }
  long long int dp[n + 1][k + 1];
  for (long long int i = 1; i < (n + 1); ++i) {
    for (long long int j = 1; j < (k + 1); ++j) {
      if (j == 1) {
        dp[i][j] = 1;
      } else if (i == 1)
        dp[i][j] = 1;
      else {
        dp[i][j] = 0;
        for (auto l : a[i]) {
          dp[i][j] = (dp[i][j] + dp[l][j - 1]) % 1000000007;
        }
      }
    }
  }
  long long int ans = 0;
  for (long long int i = 1; i < (n + 1); ++i) {
    ans = (ans + dp[i][k]) % 1000000007;
  }
  cout << ans << "\n";
  return 0;
}