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 long long mod = 1e9 + 7;
vector<long long> g[2010];
long long dp[2010][2010], sum[2010][2010];
int main() {
  long long n, k;
  cin >> n >> k;
  for (long long i = 2; i <= n; i++) {
    g[i].push_back(1);
    for (long long j = 2; j <= (int)sqrt(i); j++)
      if (i % j == 0) {
        g[i].push_back(j);
        if (i / j != j) g[i].push_back(i / j);
      }
  }
  for (long long i = 1; i <= k; i++) dp[1][i] = 1, sum[1][i] = i;
  for (long long i = 2; i <= n; i++) {
    for (long long l = 1; l <= k; l++) {
      for (long long j = 0; j < g[i].size(); j++)
        dp[i][l] += sum[g[i][j]][l - 1], dp[i][l] %= mod;
      dp[i][l]++;
      sum[i][l] = sum[i][l - 1] + dp[i][l];
      sum[i][l] %= mod;
    }
  }
  long long res = 0;
  for (long long i = 1; i <= n; i++) res += dp[i][k];
  res %= mod;
  cout << res;
}