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;
inline int gi() {
  int f = 1, x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return f * x;
}
inline long long gl() {
  long long f = 1, x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return f * x;
}
const int INF = 0x3f3f3f3f, N = 2003, M = N << 1, mod = 1000000007;
int n, m;
int dp[N][N];
vector<int> v[N];
int main() {
  n = gi(), m = gi();
  for (int i = 1; i <= n; i += 1)
    for (int j = 1; j * j <= i; j += 1)
      if (i % j == 0) {
        v[i].push_back(j);
        if (j * j != i) v[i].push_back(i / j);
      }
  dp[0][1] = 1;
  for (int i = 1; i <= m; i += 1)
    for (int j = 1; j <= n; j += 1) {
      for (auto x : v[j]) (dp[i][j] += dp[i - 1][x]) %= mod;
    }
  int ans = 0;
  for (int i = 1; i <= n; i += 1) (ans += dp[m][i]) %= mod;
  printf("%d\n", ans);
  return 0;
}