Andy, 8, has a dream - he wants to produce his very own dictionary. This is not an easy task for him, as the number of words that he knows is, well, not quite enough. Instead of thinking up all the words himself, he has a briliant idea. From his bookshelf he would pick one of his favourite story books, from which he would copy out all the distinct words. By arranging the words in alphabetical order, he is done! Of course, it is a really time-consuming job, and this is where a computer program is helpful.

You are asked to write a program that lists all the different words in the input text. In this problem, a word is defined as a consecutive sequence of alphabets, in upper and/or lower case. Words with only one letter are also to be considered. Furthermore, your program must be CaSe InSeNsItIvE. For example, words like "Apple", "apple" or "APPLE" must be considered the same.

Input

The input file is a text with no more than 5000 lines. An input line has at most 200 characters. Input is terminated by EOF.

Your output should give a list of different words that appears in the input text, one in a line. The words should all be in lower case, sorted in alphabetical order. You can be sure that he number of distinct words in the text does not exceed 5000.

Adventures in Disneyland

Two blondes were going to Disneyland when they came to a fork in the
road. The sign read: "Disneyland Left."

So they went home.

a
adventures
blondes
came
disneyland
fork
going
home
in
left
read
road
sign
so
the
they
to
two
went
were
when

題意：按字典序輸出所有出現的單詞，重複的不算。

思路：首先RE的看下，題目說輸出的單詞不超過5000個，但是沒說加上重複的有多少，我已經試過把陣列開到最大還是RE，所以這道題要用vector。另外優化下字母大小寫的轉換。對比了一下網上其他的程式碼，感覺我的程式碼是最短的，也是最容易理解的。【我沒有吹牛，真心是這麼認為的

AC程式碼如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
char c,s[10010];
vector<string> dis;
int main()
{ int i,pos=0;
  while(~(c=getchar()))
  { if(isalpha(c))
     s[pos++]=tolower(c);
    else if(pos!=0)
    { s[pos]='\0';
      dis.push_back(s);
      pos=0;
    }
  }
  sort(dis.begin(),dis.end());
  for(i=0;i<dis.size();i++)
   if(i==0 || dis[i]!=dis[i-1] )
   cout<<dis[i]<<endl;
}