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 <class T>
inline void smax(T &x, T y) {
  x = max((x), (y));
}
template <class T>
inline void smin(T &x, T y) {
  x = min((x), (y));
}
const int MOD = 1e9 + 7;
const int SIZE = 2 * 1e3 + 10;
const int INF = 0x3f3f3f3f;
int n, k;
long long dp[SIZE][SIZE];
int mod(long long x) {
  while (x < 0) x += MOD;
  return x % MOD;
}
int main() {
  if (0) {
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
  }
  while (scanf("%d%d", &(n), &(k)) == 2) {
    memset((dp), 0, sizeof((dp)));
    for (int i = 0; i < (n + 1); ++i) {
      dp[1][i] = 1;
    }
    for (int i = (2); i < (k + 1); ++i)
      for (int j = (1); j < (n + 1); ++j) {
        int k = j;
        while (k <= n) {
          dp[i][j] = mod(dp[i][j] + dp[i - 1][k]);
          k += j;
        }
      }
    long long ans = 0;
    for (int i = (1); i < (n + 1); ++i) ans = mod(ans + dp[k][i]);
    printf("%I64d\n", ans);
  }
  return 0;
}