The stable marriage problem consists of matching members of two different sets according to the member's preferences for the other set's members. The input for our problem consists of:

• a set M of n males;
• a set F of n females;
• for each male and female we have a list of all the members of the opposite gender in order of preference (from the most preferable to the least).

A marriage is a one-to-one mapping between males and females. A marriage is called stable, if there is no pair (m, f) such that f ∈ F prefers m ∈ M to her current partner and m prefers f over his current partner. The stable marriage A is called male-optimal if there is no other stable marriage B, where any male matches a female he prefers more than the one assigned in A.

Given preferable lists of males and females, you must find the male-optimal stable marriage.

Input

The first line gives you the number of tests. The first line of each test case contains integer n (0 < n < 27). Next line describes n male and n female names. Male name is a lowercase letter, female name is an upper-case letter. Then go n lines, that describe preferable lists for males. Next n lines describe preferable lists for females.

Output

For each test case find and print the pairs of the stable marriage, which is male-optimal. The pairs in each test case must be printed in lexicographical order of their male names as shown in sample output. Output an empty line between test cases.

Sample Input

2
3
a b c A B C
a:BAC
b:BAC
c:ACB
A:acb
B:bac
C:cab
3
a b c A B C
a:ABC
b:ABC
c:BCA
A:bac
B:acb
C:abc

Sample Output

a A
b B
c C

a B
b A
c C

婚姻匹配演算法：簡單說就是讓男生先選最優的，女生都不選，兩個男生選同一個，女生在選一個更合適的，依次遞推

比較坑的地方是 poj 最後一個樣例輸出或不輸出空格都能過 但 zoj 最後一個樣例不能輸出空格

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
typedef long long ll;
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define push_back pb
const int N=30+10;
int ml[N][N],wl[N][N];
int c1[N],c2[N];
int id1[N],id2[N];
int n;
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		queue<int > q;
		char name[2],str[30];
		scanf("%d",&n);	
		for(int i=1;i<=n;i++)
		{
			scanf("%s",name);
			id1[i]=name[0]-'a'+1;
			q.push(id1[i]);
		//	cout<<id1[i]<<endl;
		} 
		for(int i=1;i<=n;i++)
		{
			scanf("%s",name);
			id2[i]=name[0]-'A'+1;
		}
		sort(id1+1,id1+1+n);
		for(int i=1;i<=n;i++)
		{
			c1[i]=1;
			scanf("%s",str);
			for(int j=1;j<=n;j++)
				ml[i][j]=str[j+1]-'A'+1;
		//	for(int j=1;j<=n;j++)
		//		cout<<ml[i][j]<<endl;
		}
		for(int i=1;i<=n;i++)
		{
			scanf("%s",str);
			wl[i][0]=0;
			for(int j=1;j<=n;j++)
				wl[i][str[j+1]-'a'+1]=n+1-j;
		//	for(int j=1;j<=n;j++)
			//	cout<<wl[i][j]<<endl; 
		}
		mem(c2,0);
		while(!q.empty())
		{
			int now=q.front();
		//	cout<<now;
			int wife=ml[now][c1[now]];
		//	cout<<" "<<wife<<" "<<c2[wife]<<endl;
			if(wl[wife][now]>wl[wife][c2[wife]])
			{
				q.pop();
				
				if(c2[wife])
				{
					q.push(c2[wife]);
					c1[c2[wife]]++;
				}
				c2[wife]=now;
			}
			else c1[now]++;
		}
		for(int i=1;i<=n;i++)
			printf("%c %c\n",id1[i]-1+'a',ml[id1[i]][c1[id1[i]]]-1+'A');
		if(T)puts(""); // 坑點
	}
	return 0;
}