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.
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  n = n * 2;
  int a[1000000];
  for (int i = 0; i < n; i++) cin >> a[i];
  int sum_first = 0;
  int sum_last = 0;
  for (int i = 0; i < n / 2; i++) sum_first += a[i];
  for (int i = n / 2; i < n; i++) sum_last += a[i];
  int temp_first = sum_first;
  int temp_last = sum_last;
  int count = 0;
  if (sum_first == sum_last) {
    for (int i = 0; i < n / 2; i++) {
      for (int j = n / 2; j < n; j++) {
        if (temp_first - a[i] != temp_last - a[j] && count != 1) {
          swap(a[i], a[j]);
          for (int i = 0; i < n; i++) cout << a[i] << " ";
          cout << endl;
          count = 1;
          break;
        }
      }
    }
    if (count == 0) {
      cout << "-1" << endl;
    }
  } else {
    for (int i = 0; i < n; i++) cout << a[i] << " ";
    cout << endl;
  }
  return 0;
}
