Codeforces 414B Mashmokh and ACM【预处理+dp】

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).

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].

题目大意：

给你两个数N,K，让你找有多少种合法的序列，使得序列中元素是从1-N的，并且长度为K。

一个序列由：a1，a2，a3.........ak组成，其中需要满足这样一个要求：ai+1%ai==0;

思路：

1、经典的计数问题，设定dp【i】【j】表示长度为i的序列，其最后一个数字为j的可行方案数。

那么不难推出其状态转移方程：
dp【i】【j】=dp【i】【j】+dp【i-1】【k】（其中要求j%k==0），那么如果我们暴力求解：一层for枚举i，一层for枚举j，一层for枚举k的话时间复杂度O（n^3）,显然会超时，那么我们预处理数字J的因字数，暴力求即可时间复杂度O(N^2),设定yinzi【i】【j】表示数字i的第j个因子数，同时用cont【i】表示第i个数一共有多少个因子数。

那么我们在求dp【i】【j】的时候的第三层for枚举的时候，直接枚举cont【j】个数即可。时间复杂度优化至O（n^2*cont[i]）,明显2000以内的数的因字数不会很多，所以这个时候我们预处理的优化度就显现出来了。

2、那么我们预处理dp【i】【j】之后，其解为：

Σ（dp【k】【i】）【1<=i<=n】

Ac代码：

#include
#include
using namespace std;
#define ll __int64
#define mod 1000000007
int dp[3520][3520];
int yinzi[3520][3520];
int cont[3520];
void initinit()
{
    memset(cont,0,sizeof(cont));
    for(int i=1;i<=3505;i++)
    {
        for(int j=1;j<=i;j++)
        {
            if(i%j==0)
            {
                yinzi[i][cont[i]++]=j;
            }
        }
    }
}
void init()
{
    initinit();
    memset(dp,0,sizeof(dp));
    for(int i=1;i<=3500;i++)dp[1][i]=1;
    for(int i=2;i<=3500;i++)
    {
        for(int j=1;j<=3500;j++)
        {
            for(int k=0;k