#include <fstream>
#include <cstring>
using namespace std;
const int N = 30005;
const int M = 18;
int L[N], n, cost[M][M], d[M][(1 << M) + 3], sol[M][(1 << M) + 3], ok[M];
char s[M][N];

inline void prefix(int j) {
    int i, n, k;
    n = strlen(s[j]);
    L[1] = 0;
    k = 0;
    for(i = 2; i <= n; ++i) {
        k = L[i - 1];
        while(k && s[j][k] != s[j][i - 1])
            k = L[k];
        if(s[j][k] == s[j][i - 1])
            ++k;
        L[i] = k;
    }
}

inline int kmp(int p, int t){
    int max = 0, i, k, n = strlen(s[t]), m = strlen(s[p]);
    k = 0;
    for(i = 1; i <= n; ++i){
        while(k && s[p][k] != s[t][i - 1])
            k = L[k];
        if(s[p][k] == s[t][i - 1])
            ++k;
        if(k == m){
            return k;
            k = L[k];
        }
    }

    return k;
}

int main() {
    ifstream fin("adn.in");
    ofstream fout("adn.out");
    int i, j, sw, k;
    fin >> n;
    for(i = 0; i < n; ++i)
        fin >> s[i];

    for(i = 0; i < n; ++i){
        memset(L, 0, sizeof(L));
        prefix(i);
        for(j = 0; j < n; ++j)
            if(i != j){
                cost[j][i] = kmp(i, j);
                if(cost[j][i] == strlen(s[i])){
                    ok[i] = 1; cost[j][i] = -1;
                    break;
                }
            }
    }

    for(j = 1; j < (1 << n); ++j)
        for(i = 0; i < n; ++i)
            if(((1 << i) & j) && !ok[i])
                for(k = 0; k < n; ++k)
                    if(((1 << k) & j) && i != k && !ok[k])
                        if(d[i][j] < d[k][j ^ (1 << i)] + cost[i][k]){
                            d[i][j] = d[k][j ^ (1 << i)] + cost[i][k];
                            sol[i][j] = k;
                        }
    int max = 0, p = -1;
    for(i = 0; i < n; ++i)
        if(max < d[i][(1 << n) - 1])
            max = d[i][(1 << n) - 1], p = i;

    int solutie[M + 4], nr = 0;
    j = (1 << n) - 1;
    for(i = 0; i < n; ++i)
        if(ok[i])
            j -= (1 << i);
        while(j) {
            solutie[++nr] = p;
            i = p;
            p = sol[p][j];
            j ^= 1 << i;
        }
        fout << s[solutie[1]];
        for(i = 2; i <= nr; ++i)
            if(s[solutie[i]] + cost[solutie[i - 1]][solutie[i]] != NULL)
                fout << s[solutie[i]] + cost[solutie[i - 1]][solutie[i]];

        fout << '\n';

    return 0;
}