CodeForces 452A Eevee

G - Eevee

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 452A

Appoint description: System Crawler  (Jul 18, 2016 5:05:54 PM)

Description

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 Input

Input

7j......

Output

jolteon

Input

7...feon

Output

leafeon

Input

7.l.r.o.

Output

flareon

Hint

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"}

#include<stdio.h>#include<string.h>#include<algorithm>#include<math.h>using namespace std;int main(){char s[][50]={"vaporeon", "jolteon","flareon" , "espeon","umbreon", "leafeon", "glaceon", "sylveon"};int len[]={8,7,7,6,7,7,7,7};int n,i,j,k,l,m,flag,ans;char str[50];while(scanf("%d",&n)!=EOF){scanf("%s",str);for(i=0;i<8;i++){if(len[i]==n){flag=0;    for(j=0;j<n;j++)    {   if(str[j]!='.'&&str[j]!=s[i][j])   {  flag=1;  break;   }    }    if(flag==0)    ans=i;}}printf("%s\n",s[ans]);}return 0;}