CF---（452）A. Eevee

A. Eevee

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are solving the crossword problem K from IPSC 2014. You solved all the clues except for one: who does Eevee evolve into? You are not very into pokemons, but quick googling helped you find out, that Eevee can evolve into eight different pokemons: Vaporeon, Jolteon, Flareon, Espeon, Umbreon, Leafeon, Glaceon, and Sylveon.

You know the length of the word in the crossword, and you already know some letters. Designers of the crossword made sure that the answer is unambiguous, so you can assume that exactly one pokemon out of the 8 that Eevee evolves into fits the length and the letters given. Your task is to find it.

Input

First line contains an integer n (6 ≤ n ≤ 8) – the length of the string.

Next line contains a string consisting of n characters, each of which is either a lower case english letter (indicating a known letter) or a dot character (indicating an empty cell in the crossword).

Output

Print a name of the pokemon that Eevee can evolve into that matches the pattern in the input. Use lower case letters only to print the name (in particular, do not capitalize the first letter).

Sample test(s)

Input

7 j......

Output

jolteon

Input

7 ...feon

Output

leafeon

Input

7 .l.r.o.

Output

flareon

Note

Here's a set of names in a form you can paste into your solution:

["vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"]

{"vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"}

给你一个稀疏的字符串，告诉你长度，然后要你在已知的8个字符串下进行匹配..........

代码：

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 using namespace std;
 struct node{
   int len;
   char ss[];
 };
 node str[]={
   {,"vaporeon"},
   {,"jolteon"},
   {,"flareon"},
   {, "espeon"},
   {,"umbreon"},
   {,"leafeon"},
   {,"glaceon"},
   {,"sylveon"}
 };
 int main()
 {
   int n;
   char ss[];
 //freopen("test.in","r",stdin);
   while(scanf("%d%s",&n,ss)!=EOF)
       for(int i=;i<;i++){
           if(n==str[i].len){
               bool flag=true;
             for(int j=;j<n;j++){
               if(ss[j]!='.'&&ss[j]!=str[i].ss[j]){
                  flag=false;
                    break;
                 }
             }
             if(flag){
              printf("%s\n",str[i].ss);
              break;
             }
           }
       }
   return ;
 }