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 optimize("O3")
using namespace std;
void FastInputOutput() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
void ReadFromFile(string name, string extention) {
  string fullPath = name + "." + extention;
  freopen(fullPath.c_str(), "r", stdin);
}
inline int D() {
  int t;
  scanf("%d", &t);
  return t;
}
inline long long LLD() {
  long long t;
  scanf("%lld", &t);
  return t;
}
inline char C() {
  char c;
  scanf("%c", &c);
  return c;
}
inline long long power(long long x, long long p, long long m) {
  long long res = 1;
  while (p) {
    if (p & 1) res = (res * x) % m;
    x = (x * x) % m;
    p >>= 1;
  }
  return res;
}
long double ModLog(long double base, long double x) {
  return (logl(x) / logl(base));
}
int gcd(int a, int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
const int N = 2e3 + 5;
long long p = 1e9 + 7;
int dp[N][N];
int n, k;
int solve(int curr, int cnt) {
  if (cnt == k) return 1;
  int &ret = dp[curr][cnt];
  if (~ret) return ret;
  ret = 0;
  for (int i = 1; i * curr <= n; ++i) {
    ret += solve(curr * i, cnt + 1);
    if (ret > p) ret -= p;
  }
  return ret;
}
int main() {
  memset(dp, -1, sizeof(dp));
  n = D(), k = D();
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    ans += solve(i, 1);
    if (ans > p) ans -= p;
  }
  printf("%d", (ans + p) % p);
  return 0;
}