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 int P = 1e9 + 7;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long n) {
  long long r = 1 % P;
  for (a %= P; n; a = a * a % P, n >>= 1)
    if (n & 1) r = r * a % P;
  return r;
}
long long inv(long long x) { return x <= 1 ? 1 : inv(P % x) * (P - P / x) % P; }
const int N = 2e3 + 10;
int a[N], b[N], f[N], n, m, k;
vector<int> g[N];
int dp[N][N];
int dfs(int x, int l) {
  int& ans = dp[x][l];
  if (~ans) return ans;
  if (l == 1) return ans = 1;
  ans = 0;
  for (int y : g[x]) (ans += dfs(y, l - 1)) %= P;
  return ans;
}
int main() {
  scanf("%d%d", &n, &k), memset(dp, -1, sizeof dp);
  for (int i = 1; i <= n; ++i)
    for (int j = i; j <= n; j += i) g[i].push_back(j);
  int ans = 0;
  for (int i = 1; i <= n; ++i) (ans += dfs(i, k)) %= P;
  printf("%d\n", ans);
}