#include <bits/stdc++.h>
using namespace std;
int main() {
  std::ios::sync_with_stdio(false);
  int T;
  T = 1;
  while (T--) {
    int n, i, j;
    cin >> n;
    n = n * 2;
    int ar[n + 1];
    map<int, int> mp;
    for (i = 1; i <= n; ++i) {
      cin >> ar[i];
      ++mp[ar[i]];
    }
    if (mp.size() == 1)
      cout << -1 << endl;
    else {
      sort(ar + 1, ar + n + 1);
      for (i = 1; i <= n; ++i) cout << ar[i] << " ";
      cout << endl;
    }
  }
  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() {
  std::ios::sync_with_stdio(false);
  int T;
  T = 1;
  while (T--) {
    int n, i, j;
    cin >> n;
    n = n * 2;
    int ar[n + 1];
    map<int, int> mp;
    for (i = 1; i <= n; ++i) {
      cin >> ar[i];
      ++mp[ar[i]];
    }
    if (mp.size() == 1)
      cout << -1 << endl;
    else {
      sort(ar + 1, ar + n + 1);
      for (i = 1; i <= n; ++i) cout << ar[i] << " ";
      cout << endl;
    }
  }
  return 0;
}

```