#include <bits/stdc++.h>
using namespace std;
const int N = 2000 + 5;
int main() {
  int n, nn, a[N];
  scanf("%d", &n);
  nn = 2 * n;
  for (int i = 1; i <= nn; ++i) scanf("%d", &a[i]);
  sort(a + 1, a + 1 + nn);
  bool flag = true;
  for (int i = 1; i <= n; ++i)
    if (a[i] != a[nn - i + 1]) {
      flag = false;
      break;
    }
  if (flag)
    printf("-1");
  else
    for (int i = 1; i <= nn; ++i) printf("%d ", a[i]);
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
const int N = 2000 + 5;
int main() {
  int n, nn, a[N];
  scanf("%d", &n);
  nn = 2 * n;
  for (int i = 1; i <= nn; ++i) scanf("%d", &a[i]);
  sort(a + 1, a + 1 + nn);
  bool flag = true;
  for (int i = 1; i <= n; ++i)
    if (a[i] != a[nn - i + 1]) {
      flag = false;
      break;
    }
  if (flag)
    printf("-1");
  else
    for (int i = 1; i <= nn; ++i) printf("%d ", a[i]);
}

```