A. Eevee



time limit per test



memory limit per test



input



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



n (6 ≤ n ≤ 8) – the length of the string.

n



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"}




水题,直接试


#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (8+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
char s[9][100]={"vaporeon", "jolteon", "flareon", "espeon", "umbreon", "leafeon", "glaceon", "sylveon"},st[MAXN];
int n;
int main()
{
// freopen("Eevee.in","r",stdin);
// freopen("Eevee.out","w",stdout);
scanf("%d",&n);
scanf("%s",st);
Rep(i,8)
if (strlen(s[i])==n)
{
bool b=0;
Rep(j,n)
{
if (st[j]!='.'&&st[j]!=s[i][j]) {b=1;break;}
}
if (b) continue;
printf("%s\n",s[i]);
}


return 0;
}