You're given a list of n strings a_1, _a_2, …, _a__n.
You'd like to concatenate them together in some order such that the resulting string would be lexicographically smallest.

Given the list of strings, output the lexicographically smallest concatenation.

Input

The first line contains integer n — the number of strings (1 ≤ n ≤ 5·104).

Each of the next n lines contains one string a__i (1 ≤ |a__i| ≤ 50)
consisting of only lowercase English letters. The sum of string lengths will not exceed 5·104.

Output

Print the only string a — the lexicographically smallest string concatenation.

Examples

input

4
abba
abacaba
bcd
er

output

abacabaabbabcder

input

5
x
xx
xxa
xxaa
xxaaa

output

xxaaaxxaaxxaxxx

input

3
c
cb
cba

output

cbacbc

题意：给你n个字符串，串的长度可能不同，让你把这些串连起来使得最后的字典序最小。

思路：一开始的想法是把字符串补齐，然后排个序，但是wa了。看了q神的解法，发现就是在排序的时候用字符串a+b<b+a进行排序,太亮了这方法。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
#define inf 99999999
#define pi acos(-1.0)
#define maxn 50050
#define MOD 1000000007
using namespace std;
typedef long long ll;
typedef long double ldb;

struct node{
    char s[60];
}a[maxn];
bool cmp(node a,node b){
    char s1[120],s2[120];
    char s3[120],s4[120];
    strcpy(s1,a.s);
    strcpy(s2,b.s);
    strcpy(s3,b.s);
    strcpy(s4,a.s);
    strcat(s1,s2);
    strcat(s3,s4);
    return strcmp(s1,s3)<0;

}
int main()
{
    int n,m,i,j;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++){
            scanf("%s",a[i].s);
        }
        sort(a+1,a+1+n,cmp);
        for(i=1;i<=n;i++){
            cout<<a[i].s;
        }
        cout<<endl;
    }
    return 0;
}