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 double pi = 3.1415926535898;
const char spc = ' ';
long long binpow(long long a, long long n) {
  if (n == 0) {
    return 1;
  }
  long long res = 1;
  while (n) {
    if (n % 2 == 1) {
      res *= a;
    }
    a *= a;
    n /= 2;
  }
  return res;
}
int gcd(int a, int b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
int siz(long long x) {
  if (x < 1e1) {
    return 1;
  } else if (x < 1e2) {
    return 2;
  } else if (x < 1e3) {
    return 3;
  } else if (x < 1e4) {
    return 4;
  } else if (x < 1e5) {
    return 5;
  } else if (x < 1e6) {
    return 6;
  } else if (x < 1e7) {
    return 7;
  } else if (x < 1e8) {
    return 8;
  } else if (x < 1e9) {
    return 9;
  } else if (x < 1e10) {
    return 10;
  }
}
int inf_int = 2e9;
long long inf_ll = 2e18;
const double EPS = 1e-8;
const long long MOD = 1e9 + 7;
int t;
int n, k;
long long dp[2001][2001];
vector<vector<int> > divisors;
void solve() {
  cin >> n >> k;
  {
    vector<int> help;
    divisors.assign(n + 1, help);
  }
  for (int i = 0; i <= 2000; i++) {
    for (int j = 0; j <= 2000; j++) {
      dp[i][j] = 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= i; j++) {
      if (i % j == 0) {
        divisors[i].push_back(j);
      }
    }
  }
  for (int i = 1, j = 1; j <= n; j++) {
    dp[i][j] = 1;
  }
  for (int i = 2; i <= k; i++) {
    for (int j = 1; j <= n; j++) {
      for (int x : divisors[j]) {
        dp[i][j] += dp[i - 1][x];
        dp[i][j] %= MOD;
      }
    }
  }
  long long ans = 0;
  for (int i = k, j = 1; j <= n; j++) {
    ans += dp[i][j];
    ans %= MOD;
  }
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}