#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2")
int32_t main() {
  int n;
  cin >> n;
  int a[2 * n];
  for (__typeof(2 * n) i = (0) - ((0) > (2 * n));
       i != (2 * n) - ((0) > (2 * n)); i += 1 - (((0) > (2 * n)) << 1))
    cin >> a[i];
  sort(a, a + 2 * n);
  int s1 = accumulate(a, a + n, 0);
  int s2 = accumulate(a + n, a + 2 * n, 0);
  if (s1 != s2)
    for (__typeof(n * 2) i = (0) - ((0) > (n * 2));
         i != (n * 2) - ((0) > (n * 2)); i += 1 - (((0) > (n * 2)) << 1))
      cout << a[i] << " ";
  else
    cout << -1;
  cout << "\n";
  return 0;
}

### 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;
const int INF = 0x3f3f3f3f;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2")
int32_t main() {
  int n;
  cin >> n;
  int a[2 * n];
  for (__typeof(2 * n) i = (0) - ((0) > (2 * n));
       i != (2 * n) - ((0) > (2 * n)); i += 1 - (((0) > (2 * n)) << 1))
    cin >> a[i];
  sort(a, a + 2 * n);
  int s1 = accumulate(a, a + n, 0);
  int s2 = accumulate(a + n, a + 2 * n, 0);
  if (s1 != s2)
    for (__typeof(n * 2) i = (0) - ((0) > (n * 2));
         i != (n * 2) - ((0) > (n * 2)); i += 1 - (((0) > (n * 2)) << 1))
      cout << a[i] << " ";
  else
    cout << -1;
  cout << "\n";
  return 0;
}

```