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>
template <typename T>
T in() {
  char ch;
  T n = 0;
  bool ng = false;
  while (1) {
    ch = getchar();
    if (ch == '-') {
      ng = true;
      ch = getchar();
      break;
    }
    if (ch >= '0' && ch <= '9') break;
  }
  while (1) {
    n = n * 10 + (ch - '0');
    ch = getchar();
    if (ch < '0' || ch > '9') break;
  }
  return (ng ? -n : n);
}
template <typename T>
inline T Bigmod(T b, T p, T m) {
  if (p == 0)
    return 1;
  else if (!(p & 1))
    return (Bigmod(b, p / 2, m) * Bigmod(b, p / 2, m)) % m;
  else
    return ((b % m) * Bigmod(b, p - 1, m)) % m;
}
template <typename T>
inline T ABS(T a) {
  if (a < 0)
    return -a;
  else
    return a;
}
template <typename T>
inline T Dis(T x1, T y1, T x2, T y2) {
  return sqrt((x1 - x2 * x1 - x2) + (y1 - y2 * y1 - y2));
}
template <typename T>
inline T gcd(T a, T b) {
  if (a < 0) return gcd(-a, b);
  if (b < 0) return gcd(a, -b);
  return (b == 0) ? a : gcd(b, a % b);
}
template <typename T>
inline T lcm(T a, T b) {
  if (a < 0) return lcm(-a, b);
  if (b < 0) return lcm(a, -b);
  return a * (b / gcd(a, b));
}
template <typename T>
T ModInv(T b, T m) {
  return Bigmod(b, m - 2, m);
}
bool isVowel(char ch) {
  ch = toupper(ch);
  if (ch == 'A' || ch == 'U' || ch == 'I' || ch == 'O' || ch == 'E')
    return true;
  return false;
}
template <typename T>
T POW(T b, T p) {
  if (p == 0) return 1;
  if (p % 2 == 0) {
    T s = POW(b, p / 2);
    return s * s;
  }
  return b * POW(b, p - 1);
}
using namespace std;
using namespace std;
long long int n, k, dp[2009][2009], cnt;
long long int solve(int pos, int mx) {
  if (pos == k) return 1;
  long long int &res = dp[pos][mx];
  if (res != -1) return res;
  res = 0;
  int lim = mx;
  if (lim == 0) lim = 1;
  if (mx == 0) {
    for (int i = lim; i <= n; i += lim) {
      res += solve(pos + 1, i);
      res %= 1000000007;
    }
  } else {
    for (int i = lim; i <= n; i += lim) {
      if (i % mx == 0) {
        res += solve(pos + 1, i);
        res %= 1000000007;
      }
    }
  }
  return res;
}
int main() {
  cin >> n >> k;
  memset(dp, -1, sizeof(dp));
  long long int ans = solve(0, 0);
  cout << ans << endl;
  return 0;
}