#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(16);
  int t;
  t = 1;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a(2 * n);
    set<int> s;
    int c1 = 0, c2 = 0;
    for (int i = 0; i < 2 * n; i += 1) {
      cin >> a[i];
      s.insert(a[i]);
    }
    if (s.size() == 1) {
      cout << -1;
      continue;
    }
    sort(a.begin(), a.end());
    for (int i = 0; i < 2 * n; i += 1) {
      cout << a[i] << " ";
    }
  }
}

### Prompt

Generate 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() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(16);
  int t;
  t = 1;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a(2 * n);
    set<int> s;
    int c1 = 0, c2 = 0;
    for (int i = 0; i < 2 * n; i += 1) {
      cin >> a[i];
      s.insert(a[i]);
    }
    if (s.size() == 1) {
      cout << -1;
      continue;
    }
    sort(a.begin(), a.end());
    for (int i = 0; i < 2 * n; i += 1) {
      cout << a[i] << " ";
    }
  }
}

```