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 T getnum() {
  T num = 0;
  char c;
  do {
    c = getchar();
  } while (c != EOF && c == ' ' && c == '\n' && c == '\t');
  int sign = (c == '-' ? -1 : 1);
  if (c != '-') num += (c - '0');
  while ((c = getchar()) != EOF && c != '\n' && c != '\t' && c != ' ') {
    num *= 10;
    num += (c - '0');
  }
  num *= sign;
  return num;
}
inline string getstr() {
  string str;
  char k;
  while ((k = getchar()) == ' ' || k == '\n') {
    k = getchar();
    if (k == ' ' || k == '\n')
      continue;
    else
      break;
  }
  while ((k = getchar()) != EOF && k != '\n' && k != '\t' && k != '\v' &&
         k != '\0' && k != ' ')
    str.push_back(k);
  return str;
}
long long dp[2005][2005], ans[2005];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  memset(dp, 0, sizeof(dp));
  for (int i = int(1), _b = int(k); i <= _b; i++) {
    for (int j = int(1), _b = int(n); j <= _b; j++) {
      if (i == 1) dp[i][j] = 1;
      ans[i] = (ans[i] + dp[i][j]) % 1000000007;
      int h = j;
      while (h <= n) {
        dp[i + 1][h] = (dp[i + 1][h] + dp[i][j]) % 1000000007;
        h += j;
      }
    }
  }
  printf("%I64d\n", ans[k]);
  return 0;
}