A - I Wanna Be the Guy

Problem description

There is a game called "I Wanna Be the Guy", consisting of n levels. Little X and his friend Little Y are addicted to the game. Each of them wants to pass the whole game.

Little X can pass only p levels of the game. And Little Y can pass only q levels of the game. You are given the indices of levels Little X can pass and the indices of levels Little Y can pass. Will Little X and Little Y pass the whole game, if they cooperate each other?

Input

The first line contains a single integer n (1 ≤  n ≤ 100).

The next line contains an integer p (0 ≤ p ≤ n) at first, then follows p distinct integers a₁, a₂, ..., a_p (1 ≤ a_i ≤ n). These integers denote the indices of levels Little X can pass. The next line contains the levels Little Y can pass in the same format. It's assumed that levels are numbered from 1 to n.

Output

If they can pass all the levels, print "I become the guy.". If it's impossible, print "Oh, my keyboard!" (without the quotes).

Examples

Input

4
3 1 2 3
2 2 4

Output

I become the guy.

Input

4
3 1 2 3
2 2 3

Output

Oh, my keyboard!

Note

In the first sample, Little X can pass levels [1 2 3], and Little Y can pass level [2 4], so they can pass all the levels both.

In the second sample, no one can pass level 4.

解题思路：标记一下1~n出现的数字，如果都出现了，则输出"I become the guy."，否则输出"Oh, my keyboard!"，水过！

AC代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int main(){
 4     int n,p,q,x;bool flag=false,used[105];
 5     memset(used,false,sizeof(used));
 6     cin>>n>>p;
 7     for(int i=1;i<=p;++i){cin>>x;used[x]=true;}
 8     cin>>q;
 9     for(int i=1;i<=q;++i){cin>>x;used[x]=true;}
10     for(int i=1;i<=n;++i)
11         if(!used[i]){flag=true;break;}
12     if(flag)cout<<"Oh, my keyboard!"<<endl;
13     else cout<<"I become the guy."<<endl;
14     return 0;
15 }