Description

Dr lee cuts a string S into N pieces,s[1],…,s[N].

Now, Dr lee gives you these N sub-strings: s[1],…s[N]. There might be several possibilities that the string S could be. For example, if Dr. lee gives you three sub-strings {“a”,“ab”,”ac”}, the string S could be “aabac”,”aacab”,”abaac”,…

Your task is to output the lexicographically smallest S.

Input

The first line of the input is a positive integer T. T is the number of the test cases followed.

The first line of each test case is a positive integer N (1 <=N<= 8 ) which represents the number of sub-strings. After that, N lines followed. The i-th line is the i-th sub-string s[i]. Assume that the length of each sub-string is positive and less than 100.

Output

The output of each test is the lexicographically smallest S. No redundant spaces are needed.

Sample Input

1
3
a
ab
ac

Sample Output

aabac

思路：

一开始以为仅仅一个排序就可以解决问题，但是例如出现ba和b的情况就会出错，次思路不行。

考虑两个子字符串s1,s2结合，结合方式只有s1s2或者s2s1，结合之后长度一定。于是改变排序的条件，

按照两种方式结合之后再进行比较，按照这种规则排序之后只需要将string的数组按照顺序加起来就可以得到结果。

 1 #include<iostream>
 2 #include<algorithm>
 3 using namespace std;
 4 
 5 bool cmp(const string &s1,const string &s2){
 6     return s1+s2<s2+s1;
 7 }
 8 
 9 int main(){
10     int testcases;
11     cin>>testcases;
12     while(testcases--){
13         int num_of_substring;
14         cin>>num_of_substring;
15         string substrings[num_of_substring],result;
16         for(int i=0;i<num_of_substring;i++)
17             cin>>substrings[i];
18         sort(substrings,substrings+num_of_substring,cmp);
19         for(int i=0;i<num_of_substring;i++)
20             result=result+substrings[i];
21         cout<<result<<endl;
22     }
23     return 0;
24 }