序列中找子序列的dp

题目网址： http://codeforces.com/problemset/problem/414/B

Description

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 Input

Input

3 2

Output

5

Input

6 4

Output

39

Input

2 1

Output

2

Hint

In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3].

#include <stdio.h>
#include <string.h>
#include <iostream>

using namespace std ;
const int N=;
const int mod=1e9+;
int dp[N][N];        //dp[i][j]表示长度为i，最后一个元素为j的序列数。

int main()
{
    int n,k;
    while(cin>>n>>k)
    {
        memset(dp,,sizeof(dp));
        for(int i=;i<=n;i++)
        dp[][i]=;
        for(int t=;t<k;t++)
        {
            for(int i=;i<=n;i++)
            {
                for(int j=;i*j<=n;j++)
                {
                    dp[t+][i*j]+=dp[t][i];//从i=1循环到n，i*j即凡是i的倍数的dp值均加上上一行（序列长度少一）的dp[t][i]值。
                    dp[t+][i*j]%=mod;
                }
            }
        }
        int set=;
        for(int i=;i<=n;i++)
        {
            set+=dp[k][i];
            set%=mod;
        }
        cout<<set<<endl;
    }
    return  ;
}