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;
int fx[] = {1, -1, 0, 0};
int fy[] = {0, 0, 1, -1};
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
struct Sorter {
  int FS, SC;
};
bool cmp(Sorter A, Sorter B) {
  if (A.FS == B.FS) return A.SC < B.SC;
  return A.FS < B.FS;
}
long long dp[3000][3000];
long long n, m;
long long solve(long long i, long long level) {
  if (level >= m) return 1L;
  if (dp[i][level] != -1) return dp[i][level];
  long long ret = 0;
  for (long long j = i; j <= n; j += i)
    ret = (ret % 1000000007 + solve(j, level + 1) % 1000000007) % 1000000007;
  dp[i][level] = ret;
  return dp[i][level];
}
int main() {
  scanf("%I64d %I64d", &n, &m);
  long long ans = 0;
  memset(dp, -1, sizeof dp);
  for (long long i = 1; i < n + 1; i++) {
    ans = (ans % 1000000007 + solve(i, 1) % 1000000007) % 1000000007;
  }
  printf("%I64d\n", ans);
  return 0;
}