Mashmokh and ACM CodeForces - 414B 【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, ..., 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  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].

题意：在【1，n】范围内任取k个数，使得a[i+1]%a[i]==0

思路：计数问题加取模，想想就可能是DP了。

dp[i][j]为长度为l,以j为结尾的数量。 

思考：我当前的状态可以由那些状态转移而来，很明显，由长度为i-1,结尾为j约数转移而来。

那么有dp[i][j]=sigmadp[i-1][x] 其中x为j的约数， 那么预处理出所有约数，处理出dp[1][1~n]=1，递推即可

#include 

typedef long long ll;
using namespace std;

int MOD=1e9+7;
vector  vec[2005];
ll dp[2005][2005];


//dp[i][j]=sigmadp[i-1][j约数]
int main(void){
    ll n,k;
    cin >> n>>k;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=i;j++)
            if(i%j==0)  vec[i].push_back(j);
    for(int i=1;i<=n;i++)   dp[1][i]=1;
    for(int i=2;i<=k;i++){
        for(int j=1;j<=n;j++){
            for(int k=0;k