#include <bits/stdc++.h>
using namespace std;
int main() {
  long int t_n;
  cin >> t_n;
  long int ar[2 * t_n];
  long int i;
  for (i = 0; i < 2 * t_n; i++) {
    cin >> ar[i];
  }
  long int n = sizeof(ar) / sizeof(ar[0]);
  long int z = ar[0];
  int flag = 0;
  for (i = 1; i < 2 * t_n; i++) {
    if (ar[i] != z) {
      flag = 1;
      break;
    }
  }
  sort(ar, ar + n);
  if (flag == 0) {
    cout << "-1" << endl;
    return 0;
  }
  for (i = 0; i < 2 * t_n; i++) cout << ar[i] << " ";
  return 0;
}

### Prompt

Please formulate a CPP solution to the following problem:

You're given an array a of length 2n. Is it possible to reorder it in such way so that the sum of the first n elements isn't equal to the sum of the last n elements?

Input

The first line contains an integer n (1 ≤ n ≤ 1000), where 2n is the number of elements in the array a.

The second line contains 2n space-separated integers a_1, a_2, …, a_{2n} (1 ≤ a_i ≤ 10^6) — the elements of the array a.

Output

If there's no solution, print "-1" (without quotes). Otherwise, print a single line containing 2n space-separated integers. They must form a reordering of a. You are allowed to not change the order.

Examples

Input


3
1 2 2 1 3 1


Output


2 1 3 1 1 2

Input


1
1 1


Output


-1

Note

In the first example, the first n elements have sum 2+1+3=6 while the last n elements have sum 1+1+2=4. The sums aren't equal.

In the second example, there's no solution.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  long int t_n;
  cin >> t_n;
  long int ar[2 * t_n];
  long int i;
  for (i = 0; i < 2 * t_n; i++) {
    cin >> ar[i];
  }
  long int n = sizeof(ar) / sizeof(ar[0]);
  long int z = ar[0];
  int flag = 0;
  for (i = 1; i < 2 * t_n; i++) {
    if (ar[i] != z) {
      flag = 1;
      break;
    }
  }
  sort(ar, ar + n);
  if (flag == 0) {
    cout << "-1" << endl;
    return 0;
  }
  for (i = 0; i < 2 * t_n; i++) cout << ar[i] << " ";
  return 0;
}

```