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;
const int MOD = int(1e9 + 7);
int dp[2005][2005];
int main() {
  int N, K, ans, lim[2005];
  long long tmp;
  for (int i = (1); (i) <= int(2000); ++i) lim[i] = (int)floor(sqrt(i));
  while (~scanf("%d %d", &N, &K)) {
    for (int n = (1); (n) <= int(N); ++n) dp[n][1] = 1;
    for (int k = (2); (k) <= int(K); ++k)
      for (int n = (1); (n) <= int(N); ++n) {
        tmp = 0;
        for (int x = (1); (x) <= int(lim[n]); ++x)
          if (n % x == 0) tmp += dp[x][k - 1] + dp[n / x][k - 1];
        if (n == lim[n] * lim[n]) tmp -= dp[lim[n]][k - 1];
        dp[n][k] = tmp % MOD;
      }
    ans = 0;
    for (int i = (1); (i) <= int(N); ++i) ans = (ans + dp[i][K]) % MOD;
    printf("%d\n", ans);
  }
  return 0;
}