Problem

Taro has N character strings, each of which is L in length. Taro loves palindromes, so I want to make palindromes as long as possible by selecting some of the N character strings and arranging them in any order.

Find the longest palindrome that Taro can make. If there are more than one, output the smallest one in lexicographical order. If you can't make a palindrome no matter how you choose and arrange them, output a blank line.

Constraints

The input satisfies the following conditions.

* 1 ≤ N ≤ 1000
* 1 ≤ L ≤ 30
* s_i is a character string consisting of lowercase letters.

Input

The input is given in the following format.


N L
s1
s2
..
..
..
sN


The number N of strings and the length L of strings are given on the first line. From the second line that follows, N & plus; The character string s_i that Taro has is given to the first line.

Output

Output the longest palindrome with the smallest dictionary order. If you cannot make a palindrome, output a blank line.

Examples

Input

4 2
oi
io
rr
rr


Output

iorrrroi


Input

5 1
a
b
c
a
c


Output

acbca


Input

3 3
uki
uku
uke


Output

uku
#include <bits/stdc++.h>
using namespace std;
/*{{{*/  //template
#define rep(i,n) for(int i=0;i<n;i++)
constexpr int INF = numeric_limits<int>::max()/2;
constexpr long long LINF = numeric_limits<long long>::max()/3;
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define all(v) (v).begin(),(v).end()
#define sz(x) (int)(x).size()
#define debug(x) cerr<<#x<<":"<<x<<endl
#define debug2(x,y) cerr<<#x<<","<<#y":"<<x<<","<<y<<endl
//struct fin{ fin(){ cin.tie(0); ios::sync_with_stdio(false); } } fin_;
struct Double{ double d; explicit Double(double x) : d(x){} };
ostream& operator<<(ostream& os,const Double x){ os << fixed << setprecision(20) << x.d; return os; }
template<typename T> ostream& operator<<(ostream& os,const vector<T>& vec){ os << "["; for(const auto& v : vec){ os << v << ","; } os << "]"; return os; }
template<typename T,typename U> ostream& operator<<(ostream& os,const pair<T,U>& p){ os << "(" << p.first << ","<< p.second <<")"; return os; }
template<typename T> ostream& operator<<(ostream& os,const set<T>& st){ os<<"{"; for(T v:st) os<<v<<","; os <<"}"; return os; }
template<typename T,typename U> inline void chmax(T &x,U y){ if(y>x) x = y; }
template<typename T,typename U> inline void chmin(T &x,U y){ if(y<x) x = y; }
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
ll gcd(ll a,ll b){ if(b==0) return a; else return gcd(b,a%b); }
//constexpr double eps = 1e-14; 
constexpr double eps = 1e-10; 
constexpr ll mod = 1e9+7;
const int dx[]={1,0,-1,0} ,dy[] = {0,1,0,-1};
/*}}}*/
int n;
int l;
array<string,1024> s;

bool ok(const string& st){
    auto ss = st;
    reverse(all(ss));
    return st == ss;
}

bool ok(string a, string b){
    reverse(all(b));
    return a == b;
}

vector<pair<string,string>> v;
array<bool,1024> used;

int main(){
    cin >> n;
    cin >> l;
    rep(i,n) cin >> s[i];
    rep(i,n) used[i] = false;

    for(int i=0;i<n;i++){
        for(int j=i+1;j<n;j++){
            if(ok(s[i],s[j]) and !used[i] and !used[j]){
                v.pb(mp(min(s[i],s[j]),max(s[i],s[j])));
                used[i] = used[j] = true;
            }
        }
    }

    sort(all(v));

    string ss = "";
    rep(i,n){
        if(!used[i] and ok(s[i])){
            if(ss=="") ss = s[i];
            else ss = min(ss,s[i]);
        }
    }

    string ans = ss;

    for(int i=v.size()-1;i>=0;i--){
        ans = v[i].fi + ans + v[i].se;
    }
    cout << ans << endl;
}