UVA

/*
  这题本来没什么思路，后来查了题解，发现如果用map来做，会很方便简洁
  
  题解见blog：
  http://www.cnblogs.com/dwtfukgv/p/5572356.html
*/

#include <cstdio>
#include <iostream>
#include <map>
#include <string>
using namespace std;
map<char, string> morse;
map<string, string> dic;
const int INF = 0x3f3f3f3f;
typedef map<string, string>::iterator pointer;
//#define debug

int judge (string a, string b)
{
	if (a == b) return 0; //完全匹配
	if (a.size() > b.size()) swap(a, b); // 如果保证a的长度小于b，交换之后，就涵盖了增删的两种情况；最终solve函数中的mmin，变为了变化量的绝对值的最小值
	if (a == b.substr(0, a.size())) return b.size() - a.size(); //如果 b 的前边和a一样，那么就可以增加或删减字符使它们一样，返回长度差
	return INF;
}

string findword(const string& s)
{
	string ans = "";
	int mmin = INF, t;
	for (pointer i = dic.begin(); i != dic.end(); i++)
	{
		t = judge(s, i->second);
		if (!t && !mmin) 
		{
			ans += "!";
			return ans;
		}
		//两次精确匹配，加 "!" 后直接返回即可 
		//关于反向迭代器rbegin: http://blog.csdn.net/kjing/article/details/6936325
		
		if (t < mmin) ans = i->first;
		mmin = min(t, mmin);
	}
	
	if (mmin) ans += "?"; //若未能精确比配，加"?"
	return ans; 
}


int main()
{
	#ifdef debug
	freopen("E:\in.txt", "r", stdin);
	freopen("E:\out.txt", "w", stdout);
	#endif
	string s, ch;
		while (cin >> ch)
		{
			if (ch == "*") break;
			cin >> s;
			morse[ch[0]] = s; // 构造每个字母的Morse编码 
		} 
		
		while (cin >> s && s != "*")
		{
			for (int i = 0; i < s.size(); i++)
			dic[s] += morse[s[i]]; //找出所有单词的编码 
	//		cout << "test: " << dic[s] << endl;
		}
		
		while (cin >> s && s != "*")
		{
			cout << findword(s) << endl;
		}	
	#ifdef debug
	fclose(stdin);
	fclose(stdout);
	#endif
	return 0;
}