#include <stdio.h>
#include <iostream>
#include <vector>

using namespace std;

const int maxN = 18;
const int maxC = 32002;
const int maxCD = 2400000;
const int modH = 666013;
const int bzH = 19;

int N;
int dMin[ (1<<18) ][ 18 ];
int parent[ (1<<18) ][ 18 ];
char str[ maxN ][ maxC ];
int ln[ maxN ];
int power[ maxC ];

int solOrd[ maxN ];

int supp[ maxN ][ maxN ];
int lC;

long long hash1, hash2;

int powF( int bz, int exp ) {
    long long rez = 1;
    long long aux = bz;
    for ( int i = 0; i < 16; i++ ) {
        if ( exp & ( 1 << i ) )
           rez = ( rez * aux ) % modH;
        aux = ( aux * aux ) % modH;
    }
    return (int) rez;
}

bool eok( int X, int K ) {
    int R = 0;
    for ( int i = 0; i < maxN; i++ )
        if (  (1<<i) & K ) R++;
    return ( R == X );
}

int main()
{
#ifndef PC_COMPILE
    freopen("adn.in","r",stdin);
    freopen("adn.out","w",stdout);
#endif
#ifdef PC_COMPILE
	freopen("data.in","r",stdin);
	freopen("data.out","w",stdout);
#endif

	scanf("%d\n", &N );

	for ( int i = 0; i < N; i++ ) {
		scanf("%s\n", str[ i ] );
		ln[i] = strlen( str[i] );
	}
	for ( int i = 0; i < 30001; i++ )
	    power[ i ] = powF( bzH, i );
    int M =0;
    for ( int i = 0; i < N; i++ ) {
        hash1 = 0;
        for ( int j = 0; j < ln[i]; j++ )
            hash1 = ( hash1 + str[i][j]*power[ln[i]-(j+1)] ) % modH;
        int ins = 1;
        for ( int j = 0 && ins; j < N; j++ )
            if ( ln[i] <= ln[j] && i != j) {
                hash2 = 0;
                for ( int k = 0; k < ln[i]; k++ )
                    hash2 = ( hash2 + str[j][k]*power[ln[i]-(k+1)] ) % modH;
                if ( hash1 == hash2 ) {
                    ins = 0;
                    for ( int k = 0; k < ln[i]; k++ )
                        if ( str[i][k] != str[j][k] ) ins = 1;
                    if ( !ins ) break;
                }
                for ( int k = ln[i]; k < ln[j]; k++ ) {
                    hash2 = ( hash2 + 100*modH - str[j][k-ln[i]]*power[ ln[i]-1 ] ) % modH;
                    hash2 = ( hash2 * bzH + str[j][k] ) % modH;
                    if ( hash2 == hash1 ) {
                        ins = 0;
                        for ( int l = k-ln[i]+1; l <= k; l++ )
                            if ( str[j][l] != str[i][l-k+ln[i]-1] ) ins = 1;
                        if ( !ins ) break;
                    }

                }
            }
        if ( ins ) {
            solOrd[M] = i;
            M++;
        }
    }
    N = M;
    for ( int i = 0; i < N; i++ ) {
		ln[ i ] = ln[ solOrd[i] ];
		memcpy( str[i], str[ solOrd[i] ], sizeof( str[i] ) );
		dMin[(1<<i)][i] = ln[i];
    }

	for ( int i = 0; i < N; i++ )
		for ( int j = 0; j < N; j++ ) {
            hash1 = 0, hash2 = 0;

            for ( int l = 1; l <= min( ln[i], ln[j] ); l++ ) {
                hash1 = ( hash1 + str[i][ ln[i] - l ]*power[ l-1] ) % modH;
                hash2 = ( hash2*bzH + str[j][ l-1 ] ) % modH;

                if ( hash1 == hash2 )
                    supp[i][j] = l;
            }
	}

    for ( int i = 0; i < (1<<N); i++ )
        for ( int j = 0; j < N; j++ )
            if ( dMin[i][j] != 0 ) {
                //printf("[%d][%d] = %d (%d)\n", i, j, dMin[i][j], parent[i][j] );
                for ( int k = 0; k < N; k++ ) {
                    if ( ! (i&(1<<k) ) )
                    if ( dMin[ i + (1<<k) ][ k ] > dMin[i][j] + ln[k] - supp[j][k] || !dMin[i+(1<<k)][k] ) {
                        dMin[i+(1<<k)][k] = dMin[i][j]+ln[k]-supp[j][k];
                        parent[i+(1<<k)][k] = j;
                    }
                }
            }


    int sol = 0;
    int cfg = (1<<N)-1;

    for ( int i = 1; i < N; i++ )
        if ( dMin[cfg][sol] > dMin[cfg][i] )
            sol = i;
	int i = N-1;
	while ( i >= 0 ) {
	    solOrd[i] = sol;
	    sol = parent[ cfg ][ sol ];

	    cfg -= ( 1 << solOrd[i] );
	    i--;
	}
	solOrd[N] = N;
    for ( int i = 0; i < N; i++ ) {
	    for ( int j = 0; j < ln[ solOrd[i] ] - supp[ solOrd[i] ][ solOrd[i+1] ]; j++ )
            printf("%c", str[ solOrd[i] ][ j ] );
	}
	printf("\n");
    return 0;
}