#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int a, b, t, c = 0, i, j, x, y, x1, y1, z, cnt = 0, n, q, m, k;
  cin >> n;
  n *= 2;
  long long int arr[n];
  for (i = 0; i < n; i++) {
    cin >> arr[i];
    cnt += arr[i];
  }
  sort(arr, arr + n);
  for (i = 0; i < n / 2; i++) c += arr[i];
  if (c == cnt - c)
    cout << "-1";
  else
    for (i = 0; i < n; i++) cout << arr[i] << " ";
  return 0;
}

### Prompt

Create a solution in Cpp for 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 long int a, b, t, c = 0, i, j, x, y, x1, y1, z, cnt = 0, n, q, m, k;
  cin >> n;
  n *= 2;
  long long int arr[n];
  for (i = 0; i < n; i++) {
    cin >> arr[i];
    cnt += arr[i];
  }
  sort(arr, arr + n);
  for (i = 0; i < n / 2; i++) c += arr[i];
  if (c == cnt - c)
    cout << "-1";
  else
    for (i = 0; i < n; i++) cout << arr[i] << " ";
  return 0;
}

```