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 target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
long long max1(long long a, long long b, long long c) {
  long long d;
  if (a >= b && a >= c)
    d = a;
  else if (b >= c && b >= a)
    d = b;
  else if (c >= a && c >= b)
    d = c;
  return d;
}
long long dp[2001][2001], s = 0;
vector<long long> v[2001];
long long f(long long n, long long k, long long mod) {
  long long sum = 0;
  if (k == 1)
    return 1;
  else if (dp[n][k] != -1)
    return dp[n][k];
  else {
    for (int i = 0; i < v[n].size(); i++) {
      sum = (sum + f(v[n][i], k - 1, mod)) % mod;
    }
  }
  return dp[n][k] = sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k, p = 1, j, l = 1, i;
  cin >> n >> k;
  long long mod = 1e9 + 7;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= k; j++) {
      dp[i][j] = -1;
    }
  }
  for (i = 1; i <= k; i++) {
    dp[1][i] = 1;
  }
  for (i = 2; i <= n; i++) {
    p = 1;
    l = 1;
    for (j = 2; j * j <= i; j++) {
      if (i % j == 0) {
        p = 0;
        break;
      }
    }
    if (p == 1) {
      for (j = 1; j <= k; j++) {
        dp[i][j] = l % mod;
        l = (l + 1) % mod;
      }
    }
  }
  for (i = 4; i <= 2000; i++) {
    for (j = 1; j * j <= i; j++) {
      if (i % j == 0) {
        if (i == j * j)
          v[i].push_back(j);
        else if (i != j * j) {
          v[i].push_back(j);
          v[i].push_back(i / j);
        }
      }
    }
  }
  long long sum = 0;
  for (i = 1; i <= n; i++) {
    sum = (sum + f(i, k, mod)) % mod;
  }
  cout << sum;
}