A. MUH and Sticks

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Two polar bears Menshykov and Uslada from the St.Petersburg zoo and elephant Horace from the Kiev zoo got six sticks to play with and assess the animals' creativity. Menshykov, Uslada and Horace decided to make either an elephant or a bear from those sticks. They can make an animal from sticks in the following way:

• Four sticks represent the animal's legs, these sticks should have the same length.
• Two remaining sticks represent the animal's head and body. The bear's head stick must be shorter than the body stick. The elephant, however, has a long trunk, so his head stick must be as long as the body stick. Note that there are no limits on the relations between the leg sticks and the head and body sticks.

Your task is to find out which animal can be made from the given stick set. The zoo keeper wants the sticks back after the game, so they must never be broken, even bears understand it.

Input

The single line contains six space-separated integers li (1 ≤ li ≤ 9) — the lengths of the six sticks. It is guaranteed that the input is such that you cannot make both animals from the sticks.

Output

If you can make a bear from the given set, print string "Bear" (without the quotes). If you can make an elephant, print string "Elephant" (wıthout the quotes). If you can make neither a bear nor an elephant, print string "Alien" (without the quotes).

Sample test(s)

input

4 2 5 4 4 4

output

Bear

input

4 4 5 4 4 5

output

Elephant

input

1 2 3 4 5 6

output

Alien

Note

If you're out of creative ideas, see instructions below which show how to make a bear and an elephant in the first two samples. The stick of length 2 is in red, the sticks of length 4 are in green, the sticks of length 5 are in blue.

解题说明：此题判断棍子的长度，考虑是否有四根一样长的棍子，然后再判断剩下的棍子长度情况。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<string>#include<cmath>#include<cstdlib>using namespace std;int main(){int b[10]={0},i,count=0,count1=0,count2=0,x;for(i=0;i<6;i++){scanf("%d",&x);b[x]++;}for(i=0;i<10;i++){if(b[i]>=4){count1=1;b[i]=b[i]-4;i--;}else if(b[i]==2){count2=1;}else{count=0;}}if(count1==1&&count2==1){printf("Elephant\n");}else if(count1==1){printf("Bear\n");}else{printf("Alien\n");}return 0;}