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>
#pragma GCC optimize("O3")
using namespace std;
const long long inf = 1e17;
const int MAXM = 2e3 + 6;
const int MAXN = 2e2 + 4;
long long b[MAXM];
long long a[MAXM];
long long su[MAXM];
set<long long> s;
int n, k;
long long dp[MAXM][MAXM];
vector<int> v[MAXM];
long long solve(int idx, int prev) {
  if (idx == k) return 1;
  if (dp[idx][prev] != -1) return dp[idx][prev];
  long long ret = 0;
  if (prev == 0) {
    for (int i = 1; i <= n; i++) {
      ret += solve(idx + 1, i);
      if (ret >= 1000000007) ret %= 1000000007;
    }
  } else {
    for (int i = prev; i <= n; i += prev) {
      ret += solve(idx + 1, i);
      if (ret >= 1000000007) ret %= 1000000007;
    }
  }
  return dp[idx][prev] = ret;
}
int main() {
  memset(dp, -1, sizeof(dp));
  int i, j;
  scanf("%d", &n);
  scanf("%d", &k);
  printf("%lld\n", solve(0, 0));
  return 0;
}