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;
int n, k;
int dp[2005][2005];
int solve(int i, int A) {
  if (i == k) {
    return 1;
  }
  if (dp[i][A] != -1) return dp[i][A];
  if (i == 0) {
    int res = 0;
    for (int j = 1; j <= n; j++) {
      res = ((long long int)res + solve(i + 1, j)) % 1000000007;
    }
    return dp[i][A] = res;
  } else {
    int res = 0;
    for (int j = 1; (long long int)j * A <= (long long int)n; j++) {
      res = ((long long int)res + solve(i + 1, j * A)) % 1000000007;
    }
    return dp[i][A] = res;
  }
}
int main() {
  while (scanf("%d %d", &n, &k) != EOF) {
    memset(dp, -1, sizeof dp);
    printf("%d\n", solve(0, 0));
  }
  return 0;
}