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 <typename T>
inline string toString(T a) {
  ostringstream os("");
  os << a;
  return os.str();
}
template <typename T>
inline long long toLong(T a) {
  long long res;
  istringstream os(a);
  os >> res;
  return res;
}
template <typename T>
inline T S(T a) {
  return a * a;
}
template <typename T>
inline T gcd(T a, T b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
template <typename T>
inline unsigned long long bigmod(T a, T b, T m) {
  if (b == 0)
    return 1;
  else if (b % 2 == 0)
    return S(bigmod(a, b / 2, m)) % m;
  else
    return (a % m * bigmod(a, b - 1, m)) % m;
}
template <typename T>
inline vector<string> parse(T str) {
  vector<string> res;
  string s;
  istringstream os(str);
  while (os >> s) res.push_back(s);
  return res;
}
template <typename T>
inline unsigned long long dist(T A, T B) {
  unsigned long long res =
      (A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y);
  return res;
}
template <typename T>
inline double cosAngle(T a, T b, T c) {
  double res = a * a + b * b - c * c;
  res = res / (2 * a * b);
  res = acos(res);
  return res;
}
template <typename T>
inline T power(T base, int po) {
  T res = 1;
  if (base == 0) return 0;
  for (int i = (0); i < (po); i++) res *= base;
  return res;
}
template <typename T>
inline bool isOn(T mask, T pos) {
  return mask & (1 << pos);
}
template <typename T>
inline int Off(T mask, T pos) {
  return mask ^ (1 << pos);
}
template <typename T>
inline int On(T mask, T pos) {
  return mask | (1 << pos);
}
long long arr[10005];
vector<int> v[5000];
long long dp[2005][2005];
int Size[2005];
int N;
int main() {
  int i, j, k;
  int n, q;
  long long x, y;
  int ks, cas;
  int kk;
  cin >> n >> kk;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= i; j++) {
      if (i % j == 0) {
        v[i].push_back(j);
      }
    }
  }
  for (i = 0; i <= n; i++) {
    Size[i] = v[i].size();
  }
  memset(dp, 0, sizeof dp);
  for (i = 1; i <= n; i++) {
    dp[1][i] = 1;
  }
  for (i = 2; i <= kk; i++) {
    for (j = 1; j <= n; j++) {
      for (k = 0; k < Size[j]; k++) {
        dp[i][j] = (dp[i][j] + dp[i - 1][v[j][k]]) % 1000000007;
      }
    }
  }
  long long res = 0;
  for (i = 1; i <= n; i++) {
    res = (res + dp[kk][i]) % 1000000007;
  }
  cout << res << endl;
  return 0;
}