#include <bits/stdc++.h>
using namespace std;
int n, arr[10000], arr2[10000], mx1, mx2, mn1 = 99999999, mn2 = 99999999, sum1,
                                          sum2;
int main() {
  cin >> n;
  for (int i = 0; i < 2 * n; i++) {
    cin >> arr[i];
  }
  sort(arr, arr + 2 * n);
  for (int i = 0; i < n; i++) {
    sum1 += arr[i];
  }
  for (int i = n; i < 2 * n; i++) {
    sum2 += arr[i];
  }
  if (sum1 == sum2)
    cout << -1;
  else {
    for (int i = 0; i < 2 * n; i++) {
      cout << arr[i] << " ";
    }
  }
}

### Prompt

In cpp, your task is to solve 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 n, arr[10000], arr2[10000], mx1, mx2, mn1 = 99999999, mn2 = 99999999, sum1,
                                          sum2;
int main() {
  cin >> n;
  for (int i = 0; i < 2 * n; i++) {
    cin >> arr[i];
  }
  sort(arr, arr + 2 * n);
  for (int i = 0; i < n; i++) {
    sum1 += arr[i];
  }
  for (int i = n; i < 2 * n; i++) {
    sum2 += arr[i];
  }
  if (sum1 == sum2)
    cout << -1;
  else {
    for (int i = 0; i < 2 * n; i++) {
      cout << arr[i] << " ";
    }
  }
}

```