DP of the sub-sequence found in the sequence

Topic website: 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 is given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer in all. So he wasn ' t able to solve them. That's why he asked the him with these tasks. One of these tasks is the following.

A sequence Of  l  integers  b _{1, b 2, ..., b l  Span class= "Tex-span" > (1≤ b 1≤ b 2≤ ... ≤ 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(^{9 + 7).}

Input

The first line of input contains, space-separated integers n, k (1≤ n, k≤2000).

Output

Output a single integer-the number of good sequences of length k modulo 1000000007(^{9 + 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 is: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3].

#include <stdio.h>#include<string.h>#include<iostream>using namespacestd;Const intn= .;Const intmod=1e9+7;intDp[n][n]; DP[I][J] Indicates the number of sequences of length I and the last element is J. intMain () {intn,k;  while(cin>>n>>k) {memset (DP,0,sizeof(DP));  for(intI=1; i<=n;i++) dp[1][i]=1;  for(intt=1; t<k;t++)        {             for(intI=1; i<=n;i++)            {                 for(intj=1; i*j<=n;j++) {dp[t+1][i*j]+=dp[t][i];//the DP value from the I=1 loop to the N,i*j, which is a multiple of I, plus the dp[t][i] value of the previous line (one less sequence length). Dp[t+1][i*j]%=MoD; }            }        }        int Set=0;  for(intI=1; i<=n;i++)        {            Set+=Dp[k][i]; Set%=MoD; } cout<<Set<<Endl; }    return 0 ;}

DP of the sub-sequence found in the sequence