#include <iostream>

#include <cstdio>

#include <string>

#include <cstring>

#include <vector>

#include <algorithm>

#define M 1050

using namespace std;

int n, top;

struct edge {

    int to, vis, id;           //to代表边的终点，id代表边的编号

};

vector<edge>w[M];

string str[M];                //原来还可以这样定义字符串数组

int ans[M];

int fa[];

int find(int x)

{

    return x == fa[x] ? x : fa[x] = find(fa[x]);

}

void fleury(int loc)

{

    for (int i = ; i<w[loc].size(); i++)

        if (!w[loc][i].vis)

        {

            w[loc][i].vis = ;

            fleury(w[loc][i].to);

            ans[top++] = w[loc][i].id;

        }

    return;

}

int main()

{

    int indegree[], outdegree[];

    int T;

    scanf("%d", &T);

    while (T--)

    {

        top = ;

        for (int i = ; i<; i++)

        {

            w[i].clear();

            outdegree[i] = indegree[i] = ;

            fa[i] = i;

        }

        scanf("%d", &n);

        int a, b;

        for (int i = ; i<n; i++)

            cin >> str[i];

        sort(str, str + n);             //根据字典序排序

        edge edg;

        int start;

        for (int i = ; i<n; i++)

        {

            a = str[i][] - 'a';

            b = str[i][str[i].size() - ] - 'a';

            indegree[a]++;

            outdegree[b]++;

            fa[find(a)] = find(b);

            edg.to = b; edg.vis = ; edg.id = i;

            w[a].push_back(edg);

            if (i == )              //重要  起点必须为初始化为第一条边的出节点（字典序最小，且满足 欧拉回路 的的要求）

                start = a;

        }

        int ss = , num = , start_num = , end_num = ;

        for (int i = ; i<; i++)

        {

            if ((indegree[i] || outdegree[i]) && find(i) == i)          //find(i)==i是用来判断整个图是否连通的，因为图存在欧拉通路的条件之一就是必须是连通图

                ss++;                                                   //结合下面的ss==1，来理解，因为如果整个图是连通的，ss就只会在当i为根节点的时候+1

            if (indegree[i] != outdegree[i])

            {

                if (outdegree[i] - indegree[i] == -)

                    start = i, start_num++;                         //这里和上面的初始化start的步骤不懂，做题的时候就是卡在了选取第一个出节点的步骤

                else if (outdegree[i] - indegree[i] == )

                    end_num++;

                num++;

            }

        }

        if ((num ==  || (num ==  && start_num ==  && end_num == )) && ss == )           //存在欧拉通路的条件

        {

            fleury(start);                 //我对start的选取不是很理解

            for (int i = top - ; i >= ; i--)             //要使输出的单词按字典序输出

            {

                if (i == )

                    cout << str[ans[i]] << endl;

                else

                    cout << str[ans[i]] << ".";

            }

        }

        else

            cout << "***" << endl;

    }

    return ;

}