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;
void swapp(int &a, int &b);
double fabss(double a);
int maxx(int a, int b);
int minn(int a, int b);
int Del_bit_1(int n);
int lowbit(int n);
int abss(int a);
const double E = 2.718281828;
const double PI = acos(-1.0);
const double ESP = 1e-6;
const int inf = 0x7fffffff;
const int N = 1e6 + 6;
const int mod = 1e9 + 7;
int dp[2005][2005];
int main() {
  ios::sync_with_stdio(false);
  int n, k, i, j, o;
  cin >> n >> k;
  for (i = 1; i <= n; ++i) dp[1][i] = 1;
  for (i = 1; i <= k - 1; ++i) {
    for (j = 1; j <= n; ++j) {
      for (o = j; o <= n; o += j) {
        dp[i + 1][o] += dp[i][j];
        dp[i + 1][o] %= mod;
      }
    }
  }
  int ans = 0;
  for (i = 1; i <= n; ++i) ans += dp[k][i], ans %= mod;
  cout << ans << endl;
  return 0;
}
int maxx(int a, int b) { return a > b ? a : b; }
void swapp(int &a, int &b) {
  a ^= b;
  b ^= a;
  a ^= b;
}
int lowbit(int n) { return n & -n; }
int Del_bit_1(int n) { return n & n - 1; }
int abss(int a) { return a > 0 ? a : -a; }
double fabss(double a) { return a > 0 ? a : -a; }
int minn(int a, int b) { return a < b ? a : b; }