B. Mashmokh and ACM(dp)

http://codeforces.com/problemset/problem/414/B

B. Mashmokh and ACM

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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, ..., b_l (1 ≤ b₁ ≤ b₂ ≤ ... ≤ b_l ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally  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).

Sample test(s)

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].

题意：1~n组成的不下降序列，求出序列长度为k的序列种数，每个序列满足序列中的后一个数都能整除前一个数。

思路：后一个数的确定只与前一个数有关，设dp[i][j]表示长度为i的序列中的最后一个数为j，则dp[i][z] = dp[i][z]+dp[i-1][j],其中z是j的倍数。

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 const int MOD=;
 int dp[][];
 int main()
 {
     int n,k;
     while(cin>>n>>k)
     {
         memset(dp,,sizeof(dp));
         for (int i = ; i <= n; i++)
             dp[][i] = ;
         for (int i = ; i <= k; i++)
         {
             for (int j = ; j <= n; j++)
             {
                 for (int z = j; z <= n; z+=j)
                 {
                     dp[i][z] = (dp[i][z]+dp[i-][j])%MOD;
                 }
             }
         }
         int ans = ;
         for (int i = ; i <= n; i++)
         {
             ans+=dp[k][i];
             ans%=MOD;
         }
         cout<<ans<<endl;
     }
     return ;
 }