//Idea:  given array with duplicates (or not)  print out duplicate numbers
#include <iostream>
#include <sstream>
#include <algorithm>
#include <vector>
using namespace std;
#define FOR(i,n) for (int i=0;i<(int)n;i++)
#define FORN(i,a,b) for (int i=a;i<=(int)b;i++)




int main()
{

	int n;
	vector<int> v,r;

	while (cin>>n && n!=0)
	{
		v.clear();
		r.clear();
		v.resize(n);
		FOR(i,n) cin>>v[i];
		sort(v.begin(),v.end());	//sort!
		int prev=0;
		FOR(i,v.size()-1)
		{
			if (v[i]==v[i+1])
			{
				//first time duplicate is detected
				//and we record it in "prev"
				prev=v[i];				
			}
			else
			{	
				//no more duplicates are detected!
				if (prev>0) //if there was a duplicate, we store it
				{
					r.push_back(prev);
					prev=0;	//and reset, so we don't add it again
				}
			}		
		}
		//check if any unrecorded duplicates got left over
		//i.e. last two numbers in array were the same
		//so we could not get to them in the loop above
		if (prev>0) //for previous duplicate
		{
			r.push_back(prev);
		}
		if (r.size()>0) 
		{
			cout<<r.front();
			FORN(i,1,r.size()-1) cout<<" "<<r[i];
		}
		else cout<<"I want more dogs with large eyes!";
		cout<<endl;
	}
	return 0;
}

//You are given an array with integers between 1 and 1,000,000. One integer is in the array twice.
//How can you determine which one? Can you think of a way to do it using little extra memory. 


//how about none?
/*
all duplicates
first, 

5 1 2 3 4 5
5 1 2 2 3 4
5 1 1 3 3 4
5 1 1 1 4 6
9
no duplicates!
2
1 3
1
*/