You're given a list of n strings a1, a2, ..., an. 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 ai (1 ≤ |ai| ≤ 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
#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int abs1(long long int a) {
  if (a < 0)
    return -1 * a;
  else
    return a;
}
long long int power(long long int base, long long int n, long long int M) {
  if (n == 0) return 1;
  if (n == 1) return base;
  long long int halfn = power(base, n / 2, M);
  if (n % 2 == 0)
    return (halfn * halfn) % M;
  else
    return (((halfn * halfn) % M) * base) % M;
}
long long int findMMI(long long int n, long long int M) {
  return power(n, M - 2, M);
}
long long int max2(long long int a, long long int b) {
  if (a > b)
    return a;
  else
    return b;
}
long long int max3(long long int a, long long int b, long long int c) {
  max2(max2(a, b), c);
}
long long int min2(long long int a, long long int b) {
  if (a < b)
    return a;
  else
    return b;
}
long long int min3(long long int a, long long int b, long long int c) {
  return min2(min2(a, b), c);
}
bool cmp(string s, string a) { return ((s + a) < (a + s)); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  vector<string> v;
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    v.push_back(s);
  }
  sort((v).begin(), (v).end(), cmp);
  for (int i = 0; i < n; i++) cout << v[i];
  return 0;
}
